इ 10वी भूगोल स्थान- विस्तार प्रश्न उत्तरे

 इ 10वी भूगोल स्थान - विस्तार प्रश्न उत्तरे 

प्र1) खालील विधाने योग्य की आयोग्य ते लिहा. आयोग्य विधाने दुरुस्त करून लिहा.

1) ब्राझील हा देश प्रामुख्याने दक्षिण गोलार्धात आहे.

Ans-योग्य

2)  भारताच्या मध्यातून मकरवृत्त गेले 

Ans-अयोग्य 

 भारताच्या मध्यातून कर्कवृत्त गेला आहे

3) ब्राझीलचा रेखावृत्तीय विस्तार भारतापेक्षा कमी आहे

Ans-अयोग्य 

ब्राझीलचा रेखावृत्तीय विस्तार भारतापेक्षा जास्त  आहे

4) ब्राझील देशाच्या उत्तर भागातून विषुववृत्त जाते.

Ans-योग्य

5) ब्राझील देशाला पॅसिफिक महासागराचा किनारा लाभला आहे.

Ans-अयोग्य 

ब्राझील देशाला अटलांटिक  महासागराचा किनारा लाभला आहे

6) भारताच्या आग्नेयस पाकिस्तान हे राष्ट्र आहे

Ans-अयोग्य

भारताच्या  वायव्यस पाकिस्तान हे राष्ट्र आहे

प्र2. थोडक्यात उत्तरे लिहा 

अ) स्वातंत्र्योत्तर काळात भारत आणि ब्राझील या देशांना कोणत्या समस्यांना तोंड द्यावे लागले ?

Ans-1)सुमारे दीड शतक भारत देश ब्रिटिशांच्या

अधिपत्याखाली होता. १९४७ साली भारताला स्वातंत्र्य

मिळाले. 

2) स्वातंत्र्योत्तर काळात पहिल्या तीस वर्षात तीन युद्धाला सामोरे जाणे अनेक भागातील दुष्काळी परिस्थितीला तोंड येणे अशा अनेक समस्या असूनही भारत हा जगातील प्रमुख विकसनशील देश आहे.

3) ब्राझील हा देश 3 शतकांपेक्षा अधिक  काळ पोर्तुगीजांच्या आधिपत्याखाली होता. ब्राझीलला 1822 साली स्वातंत्र्य मिळाले.

4) विसाव्या शतकाच्या शेवटच्या कालखंडात जागतिक वित्तीय समस्यांमधून हा देश सावरला आहे.

आ) भारत आणि ब्राझील हा या दोन्ही देशात स्थान संदर्भातील कोणत्या बाबी वेगळ्या आहेत.

Ans-1) भारताचे स्थान पृथ्वीवर उत्तर व पूर्व गोलार्धात आहे . आशिया खंडाच्या दक्षिण भागात हा देश आहे.

2) पृथ्वीवर ब्राझील देशाच्या काही भाग उत्तर गोलार्धात बहुतांश भाग दक्षिण गोलार्धात आहे.  तसेच या देशाचे स्थान पश्चिम गोलार्धात दक्षिण अमेरिका खंडाच्या उत्तर भागात आहे.

 इ) भारत व ब्राझील यांचा अक्षवृत्तीय व रेखावृत्तीय विस्तार सांगा.

Ans-

1) भारत-अक्षवृत्तीय विस्तार= भारताच्या मुख्यभूमीचा अक्षवृत्तीय विस्तार ८°४° उत्तर अक्षवृत्त ते ३७°६' उत्तर अक्षवृत्त आहे. अंदमान व निकोबार या बेटांच्या समूहातील ६°४५' अक्षावरील 'इंदिरा पॉईंट' हे भारताचे अतिदक्षिण टोक आहे. 2) भारत-रेखावृत्तीय विस्तार= भारताचा रेखावृत्तीय विस्तार ६८°७ पूर्व रेखावृत्त ते ९७°२५ पूर्व रेखावृत्त आहे. 1) ब्राझील-अक्षवृत्तीय विस्तार= ब्राझीलचा अक्षवृत्तीय विस्तार ५°१५° उत्तर अक्षवृत्त ते ३३°४५ दक्षिण अक्षवृत्त आहे. 2) ब्राझील-रेखावृत्तीय विस्तार= ब्राझीलचा रेखावृत्तीय विस्तार ३४°४५° पश्चिम रेखावृत्त ते ७३°४८ पश्चिम रेखावृत्त आहे.

प्र3. अचूक पर्याय निवडून वाक्य लिहा

(Direct Answer) 

1) भारताचे सर्वात दक्षिणेकडील टोक इंदिरा पॉईंट नावाने ओळखले जाते

2) दक्षिण अमेरिका खंडातील हे दोन देश ब्राझीलच्या सीमेलगत नाहीत चिली इक्वेडोर

3) दोन्ही देशातील राजवट प्रजासत्ताक प्रकारचे आहे

ई) खालीलपैकी कोणता आकार ब्राझीलचा किनार भाग योग्य प्रकारे दर्शवतो ?(उ, ऊ, ए all direct answer in photo)

          *only tick mark answer is.                 write.                                answer *

Use as soon as English grammar rules with examples Marathi explanation

 Use as soon as English grammar rules with examples Marathi explanation 

नियम:-1) दिलेल्या दोन वाक्यतील  प्रथम क्रिया दर्शवणारे वाक्य शोधावे.

2) वाक्याच्या सुरुवातीस As soon as घेऊन त्यानंतर प्रथम क्रियादर्शविणरे वाक्य लिहावे.

3) त्यानंतर स्वल्पविराम देऊन नंतर घडलेल्या क्रियेचे वाक्य लिहावे.

4) वाक्यात when ,then, quickly, at the same time,at the time यासारखे कालदर्शक शब्दा असल्यास ते काढावेत.

                        Examples 

1)The vaccation was over, the boys were returning to school by train. 

Ans-As soon as the vacation was over ,the boys were returning to school by train. 

2)I was returning from school, I would playfully scare. 

Ans-As soon as I was returning from school, I would playfully scare. 

3)She noticed him and lost her temper. 

Ans-As soon as the noticed him, She lost her temper .

USE Hardly / Scarcely ------when English grammar with examples

 USE Hardly / Scarcely ------when English grammar with examples 

नियम:-1) वाक्याच्या सुरुवातीस As soon as /No sooner सारखे शब्द असल्यास ते काढून त्याऐवजी Hardly /scarely  चा उपयोग करावा .

2)वाक्यात As soon as नसल्यास ज्या वाक्याची क्रिया प्रथम घडली आहे त्या वाक्याच्या सुरुवातीस Hardly /scarcely लिहावे.

3) हे वाक्य पूर्ण भूतकाळात लिहावयाचे असते म्हणून Hardly नंतर  to have चे भूतकाळी रूप (had) घेऊन त्याकडे करता घ्यावा.

4) कर्त्यानंतर लगेच  क्रियापदाचे तिसरे रूप लिहावे.

5) पहिले वाक्य संपल्यानंतर when लिहावे व नंतर पुढे घडलेल्या क्रिया ची वाक्य जसेच्या तसे लिहावे वाक्यात जर वेळ दर्शवणारे शब्द असतील तर ते काढून टाकावेत.

Examples :-

1)As soon as I reached there ,I saw a snake. 

Ans-Hardly had I reached there when I saw a snake. 

2)He called me loudly as soon as the stranger began to heat him. 

Ans-Hardly had the stranger begun to beat him  when he called me loudly. 

3)As soon as we got on to the platform, the train arrived. 

Ans-Hardly had we got on to the platform when the train arrived 

Use able to /unable to/not able to English grammar rules with examples

 Use able to /unable to/not able to English grammar rules with examples

1)can हे शक्यता दर्शविण्यासाठी वापरले जाणारे model auxiliary आहे.

Can ऐवजी am/is/are +able to वापरतात .

Can't ऐवजी am/is/are unable to वापरतात .

Exa:-

1)I can ask a question. 

Ans-I am able to ask a question. 

2)He can't come to school. 

Ans-He is not able to come to school. 

2)could ऐवजी was/were able to वापरतात .

Couldn't ऐवजी was/were unable to वापरतात .Exa:-

1)I could write a letter. 

Ans-He I was able to write a letter. 

2)I could not ask him. 

Ans-I was unable to ask him. 

std 9 th Sci part 1 periodic classification of table question and answer

 2 periodic classification of table question and answer 

Page No 28:

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:

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. Average of the first and the third atomic mass b. Properties of the eighth element similar to the first c. Positive charge on the nucleus d. Sequential change in molecular formulae  e. Concentrated mass and positive charge f.  Lightest and negatively charged particle in all the atoms	1.  Dobereiner 2.  Newlands 3.  Mendeleev 4.  Moseley 5.  Rutherford 6.  Thomson

Page No 28:

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:

Choose the correct option and rewrite the statement.
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.

Page No 28:

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).

Page No 29:

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, 8O 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 = 1
7N = 2,5
20Ca = 2,8,8,2
16S = 2,8,6
4Be = 2,2
18Ar = 2,8,8
20Ca, 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 =2,5
6C = 2,4
8O = 2,6
5B = 2,3
13Al = 2,8,3
8O 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 smallest atomic radius because 19K, 3Li, 11Na are present in 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.

h. 6C, 3Li, 9F, 7N, 8O
Electronic configuration of the following elements is:
6C = 2,4
3Li = 2,1
9F = 2,7
7N =  2,5
8O = 2,6
9F has the highest nonmetallic character because  non-metallic character of an elements increases as we move from left to right in a period of the periodic table. This happens as the tendency of an atom to gain electrons increases due to increase in nuclear charge, the valence electrons are pulled in strongly by the nucleus and it becomes easier for an atom to gain electrons.
.

Page No 29:

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)

Page No 29:

Question 6:

Write short notes.
a.  Mendeleev’s periodic law.
b.  Structure of the modern periodic table.
c.  Position of isotopes in the Mendeleev’s and the modern periodic table.

ANSWER:

a.  Mendeleev’s periodic law.

• According to Mendeleev’s Periodic Law, “Physical and chemical properties of elements are 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 coloumns. The horizontal rows are called as 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 accomodate noble gases when they were discovered.

Demerits of Mendleev'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:

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 a periodic function of their atomic numbers. Modern periodic table is based on the 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:
Isotopes: Isotopes are the atoms having 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 drawback of Mendeleev's periodic table as he could not explained 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 same place depending upon their atomic number.

Page No 29:

Question 7:

Write scientific reasons.
a. Atomic radius goes on decreasing while going from left to right in a period.
b. Metallic character goes on decreasing while going from left to right in a period.
c.  Atomic radius goes on increasing down a group.
d.  Elements belonging to the same group have the same valency.
e. The third period contains only eight elements even through the electron capacity of the third shell is 18 .

ANSWER:

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

 

Page No 29:

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

sci 1 10th std ssc board chapter 1st gravitation question and answer

 Science Part I Solutions Solutions for Class 10 Science Chapter 1 GravitOn     

Page No 14:

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
Acceleration due to gravity	Nm2/kg2	Same in the entire universe
Gravitational constant	N	Depends on height

ANSWER

I	II	III
Mass	kg	Measure of inertia
Weight	N	Zero at the centre
Acceleration due to gravity	m/s2	Depends on height
Gravitational constant	Nm2/kg2	Same in the entire universe

Page No 14:

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?

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

c.  Write the three laws given by Kepler. How did they help Newton to arrive at the inverse square law of gravity?

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

a. Difference between mass and weight of an object is as follows:

 

S. No.	Mass	Weight
1.	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.
2.	Mass is an intrinsic property of a body.	Weight is an extrinsic property of a body.
3.	Mass is the measure of inertia.	Weight is the measure of force.
4.	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.
5.	The mass of a body cannot be zero.	The weight of a body can be zero.
6.	The SI unit of mass is kilogram (kg).	Since weight is a force, its SI unit is newton (N).
7.	The mass of a body can be measured using a beam balance and a pan balance.	The weight of a body can be measured using a spring balance and a weighing machine.

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. W=mg or W∝g$W=mg \mathrm{or} W\propto g$ and since the values of acceleration due to gravity on both the planets differ, thus the weight of the object will be different for both the planets.

b.  (i) A body is said to be under free fall when no other force except the force of gravity is acting on it.
(ii) The acceleration with which an object moves towards the centre of Earth during 

5.	The mass of a body cannot be zero.	The weight of a body can be zero.
6.	The SI unit of mass is kilogram (kg).	Since weight is a force, its SI unit is newton (N).
7.	The mass of a body can be measured using a beam balance and a pan balance.	The weight of a body can be measured using a spring balance and a weighing machine.

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. W=mg or W∝g$W=mg \mathrm{or} W\propto g$ and since the values of acceleration due to gravity on both the planets differ, thus the weight of the object will be different for both the planets.

b.  (i) A body is said to be under free fall when no other force except the force of gravity is acting on it.
(ii) The acceleration with which an object moves towards the centre of Earth during its free fall is called acceleration due to gravity. It is denoted by the letter ‘g’. It is a constant for every object falling on Earth’s surface.
(iii) The minimum velocity required to project an object to escape from the Earth's gravitational pull is known as escape velocity. It is given as:
ve=2gR−−−−√$v_e=\sqrt{2gR}$
(iv) The force required to keep an object under circular motion is known as centripetal force. This force always acts towards the centre of the circular path.

c. 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:
F = mv2/ r ...(i) 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:
F ∝ v2/ r ...(ii)
Now, if the planet takes time T to complete one revolution around the Sun, then its velocity v is given as:
v = 2πr/ T  ...(iii) where, r is the radius of the circular orbit of the planet
or, v ∝ r/ T ...(iv)      [as the factor 2π is a constant]
On squaring both sides of this equation, we get:
v2 ∝ r2/ T2...(v)
On multiplying and dividing the right-hand side of this relation by r, we get:

v2∝r3T2×1r$v^2\propto \frac{r^3}{T^2}\times \frac{1}{r}$ ...(vi)
According to Kepler’s third law of planetary motion, the factor r3/ T2  is a constant. Hence, equation (vi) becomes:
v2 ∝ 1/ r...(vii)
On using equation (vii) in equation (ii), we get:
F∝1r2$F\propto \frac{1}{r^2}$
Hence, the gravitational force between the sun and a planet is inversely proportional to the square of the distance between them.

d. For vertical upward motion of the stone:
S = h
u = u
v = 0
a = -g
Let t be the time taken by the ball to reach height h. Thus, using second equation of motion, we have
​−2gh=v2−u2⇒u=2gh−−−√Now, from first equation of motion, we havev=u−gtt=ug=2hg−−√   .....(i)$$\begin{gathered} -2gh=v^2-u^2 \\ \Rightarrow u=\sqrt{2gh} \\ \mathrm{Now}, \mathrm{from} \mathrm{first} \mathrm{equation} \mathrm{of} \mathrm{motion}, \mathrm{we} \mathrm{have} \\ v=u-gt \\ t=\frac{u}{g}=\sqrt{\frac{2h}{g}} .....\left(\mathrm{i}\right) \end{gathered}$$
For vertical downward motion of the stone:
S = h
u = 0
a = 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 second equation of motion, we have
​2gh=v'2⇒v'=2gh−−−√Now, from first equation of motion, we havev'=u+gt't'=v'g=2hg−−√    .....(ii)$$\begin{gathered} 2gh=v{\text{'}}^2 \\ \Rightarrow v\text{'}=\sqrt{2gh} \\ \mathrm{Now}, \mathrm{from} \mathrm{first} \mathrm{equation} \mathrm{of} \mathrm{motion}, \mathrm{we} \mathrm{have} \\ v\text{'}=u+gt\text{'} \\ t\text{'}=\frac{v\text{'}}{g}=\sqrt{\frac{2h}{g}} .....\left(\mathrm{ii}\right) \end{gathered}$$
Hence, from (i) and (ii), we observe that the time taken by the stone to go up is same as the time taken by it to come down.

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

Page No 15:

Question 3:

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

ANSWER:

At the centre of Earth, the force due to 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 Earth.

Page No 15:

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 8–√$\sqrt{8}$ T.

ANSWER:

From Kepler's third law of planetary motion, we have
T2∝r3$T^2\propto r^3$ .....(i)
Thus, when the period of revolution of planet at a distance R from a star is T, then from (i), we have
T2∝R3$T^2\propto R^3$ .....(ii)
Now, when the distance of the planet from the star is 2R, then its period of revolution becomes
T21∝(2R)3orT21∝8R3 .....(iii)$$\begin{gathered} T_1^2\propto (2R{)}^3 \\ \mathrm{or} \\ {T}_1^2\propto 8R^3 .....(\mathrm{iii}) \end{gathered}$$
Dividing (iii) by (ii), we get
T21T2=8R3 R3⇒T1=8–√T$$\begin{gathered} \frac{T_1^2}{T^2}=\frac{8R^3 }{R^3} \\ \Rightarrow T_1=\sqrt{8}T \end{gathered}$$

Page No 15:

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 MA, 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 × 105 km. 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

ANSWER:

a. Here, u = 0 
S = 5 m
t = 5 s
From second equation of motion, we have
S=ut +12gt25 =12g(5)2g=1025=0.4 m/s2$$\begin{gathered} S=ut +\frac{1}{2}g{t}^2 \\ 5 =\frac{1}{2}g(5{)}^2 \\ g=\frac{10}{25}=0.4 \mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$
Hence, the value of g on the planet is 0.4 m/s2.

b. The acceleration due to gravity of a planet is given as
g=GMr2$g=\frac{GM}{r^2}$
For planet A: gA=GMAr2A$g_{\mathrm{A}}=\frac{G{M}_{\mathrm{A}}}{r_{\mathrm{A}}^2}$
For planet B: gB=GMBr2B$g_{\mathrm{B}}=\frac{G{M}_{\mathrm{B}}}{r_{\mathrm{B}}^2}$
Now,
 gB=12gA   (Given)or, GMBr2B=GMA2r2A⇒MB=MAr2B2r2A$$\begin{gathered} g_{\mathrm{B}}=\frac{1}{2}{g}_{\mathrm{A}} \left(\mathrm{Given}\right) \\ \mathrm{or}, \frac{G{M}_{\mathrm{B}}}{r_{\mathrm{B}}^2}=\frac{G{M}_{\mathrm{A}}}{2r_{\mathrm{A}}^2} \\ \Rightarrow M_{\mathrm{B}}=\frac{M_{\mathrm{A}}{r}_{\mathrm{B}}^2}{2r_{\mathrm{A}}^2} \end{gathered}$$
Given: rA=12rB$r_{\mathrm{A}}=\frac{1}{2}{r}_{\mathrm{B}}$
⇒MB=MAr2B2(12rB)2=2MA$\Rightarrow M_{\mathrm{B}}=\frac{M_{\mathrm{A}}{r}_{\mathrm{B}}^2}{2({\displaystyle \frac{1}{2}}{r}_{\mathrm{B}}{)}^2}=2M_{\mathrm{A}}$
Thus, the mass of planet B should be twice that of planet A.

c. Mass of the object on Earth, m = 5 kg
Weight of the object on Earth, WE = 49 N
Weight of the object on Moon,
WM=16WE⇒WM=496=8.17 N$$\begin{gathered} W_{\mathrm{M}}=\frac{1}{6}{W}_{\mathrm{E}} \\ \Rightarrow W_{\mathrm{M}}=\frac{49}{6}=8.17 \mathrm{N} \end{gathered}$$ 
Mass of the object on Moon = 5 kg  (since mass is independent of the place of observation)

d. For vertical upward motion of the object,
S = 500 m
g = −$-$10 m/s2
v = 0
Let u be the initial velocity of the object. From third equation of motion, we have
v2−u2=2aS⇒0−u2=−2×10×500⇒u=100 m/s$$\begin{gathered} v^2-u^2=2aS \\ \Rightarrow 0-u^2=-2\times 10\times 500 \\ \Rightarrow u=100 \mathrm{m}/\mathrm{s} \end{gathered}$$
Now, let t1 be time taken by the object to reach at 500 m height. Thus,
v=u+at⇒0=100−10×t1t1=10 s $$\begin{gathered} v=u+at \\ \Rightarrow 0=100-10\times t_1 \\ {t}_1=10 \mathrm{s} \end{gathered}$$
For vertical downward motion of the object,
S = 500 m
g = 10 m/s2
u = 0
Let t2 be the time taken by the object to come back to the Earth from height of 500 m.
From second equation of motion, we have
S=ut+12at2⇒500=102×t22⇒t2=10 s     $$\begin{gathered} S=ut+\frac{1}{2}a{t}^2 \\ \Rightarrow 500=\frac{10}{2}\times t_2^2 \\ \Rightarrow t_2=10 \mathrm{s} \end{gathered}$$
Thus, the total time taken by the object to reach back to Earth = t1 + t2 = 20 s

e. Here, t =1 s
g = 10 m/s2
u = 0
Let v be the velocity of the ball on reaching the ground.

S=ut+12at2⇒500=102×t22⇒t2=10 s     $$\begin{gathered} S=ut+\frac{1}{2}a{t}^2 \\ \Rightarrow 500=\frac{10}{2}\times t_2^2 \\ \Rightarrow t_2=10 \mathrm{s} \end{gathered}$$
Thus, the total time taken by the object to reach back to Earth = t1 + t2 = 20 s

e. Here, t =1 s
g = 10 m/s2
u = 0
Let v be the velocity of the ball on reaching the ground.
Thus, from first equation of motion, we have
v = u + gt
⇒$\Rightarrow$v = 10×$\times$1 = 10 m/s
Hence, the speed of the object on reaching the ground is 10 m/s.
Let h be the height of the table. Thus, from second equation of motion, we have
S=ut+12gt2⇒h=0+12×10×12⇒h=5 m$$\begin{gathered} S=ut+\frac{1}{2}g{t}^2 \\ \Rightarrow h=0+\frac{1}{2}\times 10\times 1^2 \\ \Rightarrow h=5 \mathrm{m} \end{gathered}$$
Hence, the height of the table is 5 m.

f. The gravitational force between the Moon and the Earth can be found out using the formula,
F=GMeMmR2$F=G\frac{M_{\mathrm{e}}{M}_{\mathrm{m}}}{R^2}$
where, Me and Mm are the masses of the Earth and the Moon, respectively. Using all the given values, we have
F=(6.7×10−11)(6×1024)(7.4×1022)(3.8×105×1000)2=20.6×1019≃2×1020 N$F=(6.7\times {10}^{-11})\frac{(6\times {10}^{24})(7.4\times {10}^{22})}{(3.8\times {10}^5\times 1000{)}^2}=20.6\times {10}^{19}\simeq 2\times {10}^{20} \mathrm{N}$

g. The gravitational force between the Sun and the Earth can be found out using the formula,
F=GMeMsR2$F=G\frac{M_{\mathrm{e}}{M}_{\mathrm{s}}}{R^2}$
where, Me and Ms are the masses of the Earth and the Sun, respectively. Using all the given values, we have
3.5×1022=(6.7×10−11)(6×1024)×Ms(1.5×1011)2⇒Ms=(3.5×1022)×(1.5×1011)2(6.7×10−11)×(6×1024)=1.96×1030  kg