NCERT Solutions for Class 10 Science Chapter 5 Periodic Classification of Elements

NCERT Solutions for Class 10 Science Chapter 5 Periodic Classification of Elements

Edited By Vishal kumar | Updated on Aug 11, 2023 12:28 PM IST | #CBSE Class 10th

NCERT Solutions for Class 10 Science Chapter 5 – CBSE Access Free PDF

NCERT solutions for class 10 science chapter 5 - Welcome to Careers360 class 10 science chapter 5 question answer - Periodic Classification of Elements! If you are a Class 10 student seeking to comprehend the NCERT Solution of the periodic classification of elements related to the periodic table, you are in the right place. Our expert educators have meticulously crafted these class 10 science chapter 5 exercise question answer to align perfectly with the CBSE guidelines and the NCERT textbook. With our user-friendly and accurate solutions, you can strengthen your understanding of the periodic table and enhance your performance in exams.

Periodic Classification of Elements is the 5th chapter of NCERT Class 10 science book. While studying this chapter and solving questions, students can get help from NCERT Periodic Classification of Elements Class 10 Science chapter 5 solutions for more clarification. We have provided solutions for all the questions asked in the book. Students looking for NCERT Solutions for Periodic Classification of Elements Class 10 science chapter 5 pdf download can be used to prepare offline.

This article proves to be beneficial in comprehending the class 10 science ch 5 NCERT solutions. It provides a comprehensive understanding of the exercise questions, aiding in the completion of assignments and preparation for exams.

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NCERT Solutions for Class 10 Science: Important Formulas and Diagrams + eBook link

The class 10 science chapter 5 exercise question answer provides a valuable resource for students to understand important formulas and diagrams in a concise and comprehensive manner. This eBook link offers easy access to the solutions, enabling students to reinforce their knowledge and excel in their Science studies and exams. With expertly curated content and step-by-step explanations, these NCERT class 10 science chapter 5 exercise solutions are designed to enhance learning and boost academic performance.

  • In 1817, Dobereiner arranged some elements into groups that had similar properties. He called these groups triads. When these elements were arranged in the increasing sequence of increasing atomic masses, the atomic mass of the middle element was found to be approximately equal to the average of the atomic masses of the other two elements.

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  • In the year 1866, Newlands’ also made the arrangement of elements known at that time in the order of their increasing atomic masses.

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  • The periodic table formulated by Mendeleev contained vertical columns called groups and horizontal rows called periods.

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Download Ebook - NCERT Class 10 Science: Chapterwise Important Formulas, Diagrams, And Points

NCERT Solutions for Class 10 Science Chapter 5 Periodic Classification of Elements - Important Topics

Important Topics Covered in class 10 science chapter 5 exercise question answer– Periodic Classification Of Elements:

  • Early Attempts At the Classification of Elements:

Dobereiner's Triads: Concept and Examples

Newlands' Law of Octaves: Explanation and Examples

  • Mendeleev’s Periodic Table:

Introduction to Mendeleev's Periodic Table

Achievements of Mendeleev's Classification

Limitations of Mendeleev's Periodic Table

  • The Modern Periodic Table:

Definition and Features of the Modern Periodic Table

Position of Elements in the Modern Periodic Table

Trends in the Modern Periodic Table: Valency, Atomic Size, Metallic and Non-metallic Properties

The chapter periodic classification of elements class 10 solutions explores the early attempts at element classification, such as Dobereiner's Triads and Newlands' Law of Octaves. It then goes into Mendeleev's Periodic Table, noting both its successes and weaknesses. Finally, the Modern Periodic Table is introduced, with characteristics, element placements, and trends in element attributes discussed. Understanding these fundamental concepts is critical for students to understand periodicity and element arrangement in the periodic table.

Download Class 10 science chapter 5 question answer Periodic Classification of Elements for free.

According to the latest update on the CBSE Syllabus 2023-24, the chapter "Periodic Classification Of Elements" has been removed from the Class 10 Science syllabus. Students are advised to refer to the updated CBSE NCERT syllabus to stay informed about the topics that will be covered in their examinations.

NCERT Solutions for Class 10 Science Chapter 5 Periodic Classification of Elements

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Access Answer to NCERT Class 10 Science Chapter 5: Periodic Classification of Elements

Below are NCERT class 10 science chapter 5 exercise solutions, organized by topic for better student comprehension and use:

Topic 5.1 Making order out of chaos - Early attempts at the classification of elements

Q.1. Did Döbereiner’s triads also exist in the columns of Newlands’ Octaves? Compare and find out.

Answer:

Yes, Dobereiner's triads also exist in the column of Newland's octaves.

Newland's law of octaves, says that every eight elements had properties similar to that of the first.

For example, the element Lithium, sodium, and potassium are present in the second column of Newland's classification of elements. If lithium is the first element then, the sodium will be the 8th element and considering again sodium as the 1st element, then the 8th element will be potassium.

So, that means according to Newland's Law of octaves, lithium, sodium, and potassium should have similar properties and also these elements form a Dobereiner's traid having similar chemical properties.

Q.2. What were the limitations of Döbereiner’s classification?

Answer:

The limitation of Doberenier's classification:

Döbereiner could identify only three triads from the elements known at that time. Hence, this system of classification into triads was not found to be useful.

Q.3. What were the limitations of Newlands’ Law of Octaves?

Answer :

The limitations of Newland's law of octaves are:

(1) It was found that the Law of Octaves was applicable only up to calcium, as after calcium every eighth element did not possess properties similar to that of the first.

(2) It was assumed by Newlands that only 56 elements existed in nature and no more elements would be discovered in the future. But, later on, several new elements were discovered, whose properties did not fit into the Law of Octaves.

(3) Newland put two elements together in one slot and that too in the column of unlike elements having different properties.

Example: Cobalt and Nickel were put together in just one slot and that too in the column of elements like fluorine, chlorine, and bromine which were very different properties from these elements.

CBSE Class 10 science ch 5 NCERT solutions - Periodic Classification of Elements

Topic 5.2 Making order out of chaos: Mendeléev’s Periodic Table

Q.1. (a) Use Mendeléev’s Periodic Table to predict the formulae for the oxides of the element:

  • K

Answer:

The general formula for the oxides of the element Potassium (K): Potassium belongs to the group I of Mendeleev's periodic table, Hence, the formula of its oxide will be K_{2}O.

Q.1. (b) Use Mendeléev’s Periodic Table to predict the formulae for th e oxides of the element:

  • C

Answer:

The general formula for the oxides of the element Carbon (C): Carbon belongs to the group IV of Mendeleev's periodic table,

Hence, the formula of its oxide will be CO_{2}.

Q.1. (c) Use Mendeléev’s Periodic Table to predict the formulae for the oxides of the element :

  • Al

Answer:

The general formula for the oxides of the element Aluminium (Al): Aluminum belongs to the group III of Mendeleev's periodic table,

Hence, the formula of its oxide will be Al_{2}O_{3.}

Q.1. (d) Use Mendeléev’s Periodic Table to predict the formulae for the oxides of the following elements:

  • Si

Answer:

The general formula for the oxides of the element Silicon (Si): Silicon belongs to the group IV of Mendeleev's periodic table,

Hence, the formula of its oxide will be SiO_{2.}

Q.1. (e) Use Mendeléev’s Periodic Table to predict the formulae for the oxides of the element:

  • Ba

Answer:

The general formula for the oxides of the element Barium (Ba): Barium belongs to the group II of Mendeleev's periodic table,

Hence, the formula of its oxide will be BaO.

Q.2. Besides gallium, which other elements have since been discovered that were left by Mendeléev in his Periodic Table? (any two)

Answer:

Besides gallium Scandium (Sc) and Germanium (Ge) elements have been discovered for which gaps were left by Mendeleev in his periodic table.

Q.3. What were the criteria used by Mendeléev in creating his Periodic Table?

Answer:

The criteria used by Mendeleev in creating his periodic table were:

(i) The atomic mass of the elements was used to arrange the elements.

(ii) Grouping the elements according to their similar chemical properties.

And hence proposed that the chemical properties of the elements are the periodic function of their atomic masses and thus he arranged the elements according to the increasing order of atomic masses.

Q.4. Why do you think the noble gases are placed in a separate group?

Answer:

Noble gases are inert nature elements. In the periodic table, they are placed in a separate group because they have a unique property and are very different from other elements.

NCERT solutions for class 10 science chapter 5 Periodic Classification of Elements

Topic 5.3: Making order out of chaos: The Modern Periodic Table

Q.1. How could the Modern Periodic Table remove various anomalies of Mendeléev’s Periodic Table?

Answer:

Modern periodic table removes, various anomalies of Mendeleev's periodic table as follows:

(1) In the modern periodic table, elements are arranged in the increasing order of their atomic number, not in the atomic mass.

(2) In the modern periodic table, the isotopes are placed in the same place as they having the same atomic number but different atomic masses.

(3) Now, the same group elements have the same number of valence electrons.

(4) Hydrogen is now placed in the group containing the same valence electrons.

Q.2. Name two elements you would expect to show chemical reactions similar to magnesium. What is the basis for your choice?

Answer:

Elements that show similar chemical reactions are Beryllium (Be) and Calcium (Ca) because they belong to the same group. i.e., group 2nd in the modern periodic table

Q.3. (a) Name

  • Three elements that have a single electron in their outermost shells.

Answer:

Single valence electron elements are: (a) Lithium, Sodium, and Potassium .

Q.3.(b) Name

Answer:

Magnesium and Calcium have two electrons in their outermost shells.

Q.3. (c) Name

Answer:

The elements which have fully filled outermost shells are noble gases.

Example: Helium, Neon, and Argon.

Q.4. (a) Lithium, sodium, potassium are all metals that react with water to liberate hydrogen gas. Is there any similarity in the atoms of these elements?

Answer:

Yes , they have the following similar properties:

(1) Belongs to the same group I in the periodic table.

(2) Have 1 valence electron in the valence shell.

(3) All are alkali metals.

(4) They are highly reactive.

Q.4. (b) Helium is an unreactive gas and neon is a gas of extremely low reactivity. What, if anything, do their atoms have in common?

Answer:

Both Helium and Neon are noble gases which belong to the zero groups in the modern periodic table. They have fully filled the outermost shell with electrons. Helium has two electrons in its K shell whereas Neon has 8 electrons in its outermost shell.

Q.5. In the Modern Periodic Table, which are the metals among the first ten elements?

Answer:

Among the first ten elements metals are only 2 elements:

Lithium (Li) and Beryllium (Be) .

Q.6. By considering their position in the Periodic Table, which one of the following elements would you expect to have maximum metallic characteristics?

  • Ga \: \: \: Ge\: \: \: As\: \: \: \: Se \: \: \: \: Be

Answer:

The metallic character of the elements decreases as we move horizontally from left to right in the periodic table.

Therefore, Be will have the maximum metallic character because of its extreme left position in the periodic table.

NCERT solutions for class 10 science chapter 5 Periodic classification of elements- Exercise solutions

Q.1. Which of the following statements is not a correct statement about the trends when going from left to right across the periods of Periodic Table.

(a) The elements become less metallic in nature.
(b) The number of valence electrons increases.
(c) The atoms lose their electrons more easily.
(d) The oxides become more acidic.

Answer:

When going from left to right across the periods of the periodic table, the number of valence electrons increases. So, the tendency to lose electrons decreases whereas to accept electron increases.

Hence the option (C) is incorrect.

Q.2. Element X forms a chloride with the formula XCl_{2} , which is solid with a high melting point. X would most likely be in the same group of the Periodic Table as

(a) Na \; \; \; \; \; (b) Mg \; \; \; \; \; (c) AI \: \; \; \; \; \; (d) Si

Answer :

X must be Magnesium (Mg) which has a valency of 2 and when it combines with chlorine having valency 1, it forms MgCl_{2} .

Hence, the option (B) is correct.

Q.3.(a) Which element has

  • two shells, both of which are completely filled with electrons?

Answer:

The element which has two shells, which are both completely filled with electrons is,

Neon with electronic configuration (2,8) .

Q.3.(b) Which element has

  • The electronic configuration 2, 8, 2?

Answer:

The given electronic configuration (2,8,2) is of Magnesium (Mg).

Q.3. (c) Which element has

  • A total of three shells, with four electrons in its valence shell?

Answer:

The element having three shells and four electrons in its valence shell is Silicon (Si) .

Q.3. (d) Which element has

  • A total of two shells, with three electrons in its valence shell?

Answer:

The element having two shells and three electrons in its valence shell is Boron (B) .

Q.3. (e) Which element has

  • Twice as many electrons in its second shell as in its first shell?

Answer:

The element having twice as many electrons in its second shell as in its first shell will be:

Carbon with electronic configuration: (2,4)

Q.4. (a) What property do all elements in the same column of the Periodic Table as boron have in common?

Answer:

The same column of the periodic table as boron belongs to the group 13th has 3 valence electrons,

Hence, its valency is 3.

Therefore, all the elements in the same column of the periodic table as Boron will have a valency of 3.

Q.4. (b) What property do all elements in the same column of the Periodic Table as fluorine have in common?

Answer:

Fluorine belongs to the group 17th in the periodic table and having 1 valence electron.

Hence its valency is 1.

Therefore, all elements in the same column of the periodic table as fluorine will have a valency of 1.

Q.5. An atom has electronic configurations 2, 8, and 7.

(a) What is the atomic number of this element?
(b) To which of the following elements would it be chemically similar? (Atomic numbers are given in parentheses.)

  • N(7) F(9) P(15) Ar(18)

Answer:

The electronic configuration: (2,8,7) is of Chlorine.

(a) The atomic number of this element will be: 2+8+7 = 17 .

(b) It will be chemically similar to the element Fluorine (9) as it also has the same number of valence electrons in its valence shell.

Q.6. (a) The position of three elements A, B and C in the Periodic Table are shown below–

Group 16Group 17
--
-A
--
BC
  • State whether A is a metal or non-metal.

Answer:

As we know in the periodic table, group 3 - group 12 are transition metals and group 13 to group 16 are metalloids and then there are halogens in 17 groups.

A is in group 17 so it is a non-metal.

Q.6.(b) The position of three elements A, B, and C in the Periodic Table are shown below –

Group 16Group 17
--
-A
--
BC
  • State whether C is more reactive or less reactive than A.

Answer:

Element C is less reactive than A because in the halogens the reactivity decreases as we move down. This happens because as we move down the nuclear attraction increases due to an increase in atomic mass. Stronger attraction toward electron will make it harder to react with any other element that's why they are less reactive.

Q.6.(c) The position of three elements A, B and C in the Periodic Table are shown below –

Group 16Group 17
--
-A
--
BC

Will C be larger or smaller in size than B?

Answer:

C is much smaller than the size of the B because C has a higher atomic mass which means higher attraction to the electron which means compact (less) size.

Q.6. (d) The position of three elements A, B ,and C in the Periodic Table are shown below –

Group 16Group 17
--
-A
--
BC
  • State whether A is a metal or non-metal.
  • Which type of ion, cation or anion, will be formed by element A?

Answer:

A will form an anion as it accepts an electron to complete its octet.

Q.7. Nitrogen (atomic number 7) and phosphorus (atomic number 15) belong to group 15 of the Periodic Table. Write the electronic configuration of these two elements. Which of these will be more electronegative? Why?

Answer:

The electronic configuration of the element Nitrogen (N) is (2,5) .

The electronic configuration of the element phosphorus (P) is (2,8,5)

So, Nitrogen will be more electronegative than Phosphorus because its atom has a small size due to which the attraction of its nucleus for the incoming electron is more.

Q.8. How does the electronic configuration of an element relate to its position in the Modern Periodic Table?

Answer:

The electronic configuration of an element gives the distribution of electrons in various shells present in the element. The atoms with similar electronic configuration are placed in the same column, hence groups have the same number of valence electrons.

Group number can be known by knowing the number of valence electrons.

The number of shells present in an element is equal to the period number.

Thus, the information of electronic configuration is enough to know the group and period number of an element.

Q.9. In the Modern Periodic Table, calcium (atomic number 20) is surrounded by elements with atomic numbers 12, 19, 21, and 38. Which of these have physical and chemical properties resembling calcium?

Answer:

Element with atomic no. 12 and atomic no. 38 will have chemical properties the same as calcium. this is due to the fact that they have valency the same as calcium and also lie in the same group.

Q.10. Compare and contrast the arrangement of elements in Mendeléev’s Periodic Table and the Modern Periodic Table.

Answer:

Mendeleev's periodic table Modern periodic table
Elements are arranged in the increasing order of their atomic masses Elements are arranged in the increasing order of their atomic numbers
Consists of 8 groups Consists of 18 groups
Groups are categorized into subgroups 'a' and 'b' No subdivision of groups.
Noble gases were not grouped since they were not discovered yet Noble gases were grouped together at the end of the table.
There was no place for isotopes Slots for isotopes present according to atomic number

NCERT Class 10 Science Solutions Chapter 5 - Weightage in Board Exams

The Class 10 science chapter 5 exercise solutions on Periodic Classification of Elements have been excluded from the CBSE syllabus for the academic year 2023-24, and as a result, it will not be included in the Class 10 Science board exam. Even though the chapter has been removed, it's worth noting that understanding the fundamental concepts of periodicity and the periodic table can be beneficial for students in comprehending other chapters in the syllabus, such as Chemical Bonding and Chemical Thermodynamics.

For students preparing for the ch 5 science class 10 Science exam, it is advisable to prioritize studying the other chapters that are part of the syllabus. While the chapter on Periodic Classification of Elements can be used as a reference, studying it in detail is not essential for the exam.

Some key points to remember from class 10 Science chapter 5 NCERT solutions:

key points from NCERT Class 10 Science Chapter 5, "Periodic Table":

Periodic Classification: The class 10 science ch 5 question answer introduces the concept of periodic classification of elements, which is the arrangement of elements in order of their atomic number so that elements with similar properties are placed together.

Early Attempts: science chapter 5 class 10 discusses early periodic classification attempts, such as Dobereiner's triads and Newlands' law of octaves, which established the groundwork for organising elements based on their properties.

Mendeleev's Periodic Table: The chapter explores Mendeleev's pioneering work in creating the periodic table. He arranged elements based on their increasing atomic mass and left gaps for undiscovered elements, accurately predicting their properties.

Modern Periodic Table: Students learn about the modern periodic table, which is organized based on atomic numbers. The table consists of 18 vertical columns known as groups and 7 horizontal rows known as periods.

Trends in the Periodic Table: The class 10 chapter 5 science discusses important periodic table trends such as atomic size, ionisation energy, and electronegativity, helping in understanding the periodic behaviour of elements.

Applications: The periodic table serves as a powerful tool for predicting element properties, understanding chemical reactions, and identifying elements with specific characteristics, making it an indispensable resource for scientists and researchers.

By studying this periodic table ch 5 science class 10 NCERT solutions, students gain a deeper understanding of the organization and significance of the periodic table, a fundamental concept in chemistry that forms the basis for understanding the properties and behaviour of elements.

Key Features of NCERT Solutions for Class 10 Science Chapter 5 Periodic Classification of Elements

  • These class 10 science ch 5 question answer cover all the questions mentioned in the chapter.

  • NCERT solutions for Class 10 science chapter 5 are prepared by the experts as per CBSE guidelines.

  • These NCERT Class 10 Science solutions chapter 5 are prepared in simple language. Students will get a very good explanation of the solution for each question.

  • NCERT Solutions Class 10 Science chapter 5 pdf download is not available. But students can download them as a webpage file in the browser. Alternatively, they can use the webpage to pdf tool to save the file as a pdf.

  • Going through the free Class 10 Science Chapter 5 NCERT solutions, students will be able to understand the theory behind every concept which helps them to solve the questions quickly and accurately.

Also Check NCERT Books and NCERT Syllabus here:

CBSE NCERT Solutions for Class 10 Science - Chapter wise

NCERT Science Exemplar Solutions Class 10 - Chapter Wise

Apart from NCERT class 10 science chapter 5 exercise solutions, the NCERT exemplar solutions are also important for achieving a high score. These exemplar solutions are listed below in a chapter-wise format:

If you have any doubts about class 10 science ch 5 question answer or unable to answer them correctly then, don't worry NCERT Class 10 Science solutions chapter 5 are for you. For maths subject, you should visit NCERT class 10 Mathematics solutions.

Also, check


Frequently Asked Questions (FAQs)

1. What are the important topics of Class 10 Science Chapter 5 periodic classification of elements ?

Here is the list of important topics

  •  Variation in Properties

  •  Valency

  •  Atomic number

  •  Metallic and Non-metallic properties.

2. What is the weightage of chapter 5 periodic classification of elements in CBSE class 10 board final exam?

Questions of 5-6 marks can be asked from NCERT solutions for class 10 Science chapter 5.

3. position of hydrogen in periodic table

Hydrogen is placed in Group 1 of the periodic table but exhibits unique properties, making its position distinct and significant

4. What is the chapter number of the periodic classification of elements class 10?

The chapter number of "Periodic Classification of Elements" in Class 10 Science is Chapter 5.

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Hello Aspirant,  Hope your doing great,  your question was incomplete and regarding  what exam your asking.

Yes, scoring above 80% in ICSE Class 10 exams typically meets the requirements to get into the Commerce stream in Class 11th under the CBSE board . Admission criteria can vary between schools, so it is advisable to check the specific requirements of the intended CBSE school. Generally, a good academic record with a score above 80% in ICSE 10th result is considered strong for such transitions.

hello Zaid,

Yes, you can apply for 12th grade as a private candidate .You will need to follow the registration process and fulfill the eligibility criteria set by CBSE for private candidates.If you haven't given the 11th grade exam ,you would be able to appear for the 12th exam directly without having passed 11th grade. you will need to give certain tests in the school you are getting addmission to prove your eligibilty.

best of luck!

According to cbse norms candidates who have completed class 10th, class 11th, have a gap year or have failed class 12th can appear for admission in 12th class.for admission in cbse board you need to clear your 11th class first and you must have studied from CBSE board or any other recognized and equivalent board/school.

You are not eligible for cbse board but you can still do 12th from nios which allow candidates to take admission in 12th class as a private student without completing 11th.

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A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

Option 2)

200 \, \, J - 500 \, \, J

Option 3)

2\times 10^{5}J-3\times 10^{5}J

Option 4)

20,000 \, \, J - 50,000 \, \, J

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

Option 2)

increase two fold

Option 3)

remain unchanged

Option 4)

be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1)

Molality

Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

Option 2)

6.023 × 1022

Option 3)

half that in 8 g He

Option 4)

558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

Option 2)

more than 3 but less than 6

Option 3)

more than 6 but less than 9

Option 4)

more than 9

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