NCERT Exemplar Class 11 Maths Solutions Chapter 1 Sets

# NCERT Exemplar Class 11 Maths Solutions Chapter 1 Sets

Edited By Ravindra Pindel | Updated on Sep 12, 2022 04:38 PM IST

NCERT Exemplar Class 11 Maths solutions chapter 1 Sets deals with a variety of definitions and terms related to a set which was first introduced by Georg Cantor which serves a very important part at each, and every branch of mathematics in present times. The chapter has its reach in various different topics such as geometry, trigonometry, sequence, etc., which makes it even more vital to be understood. With the help of NCERT Exemplar Class 11 Maths chapter 1 solutions, students will be able to understand the application of sets. NCERT Exemplar solutions Class 11 Maths chapter 1 also helps in identifying whether a collection could be termed as a set or not. The essential rules like a set is denoted by a capital letter, its members or objects are represented by small letters, and the terms and objects of the collection should be synonymous, are also explained.

## NCERT Exemplar Class 11 Maths Solutions Chapter 1: Short Answer Type

Question:

1. Write the following sets in the roaster form:

is a positive factor of a prime number

(i) Given that

; Hence ,

(ii) Given that

, ; Hence

(iii) Given that is a positive factor of a prime number

So , the positive factors of prime number P are 1 and P . Hence ,

Question:

Write the following sets in the roaster form:

(i) Given that

(ii) Given that

Hence ,
(iii) Given that

Hence

Question:

Write the following sets in the roaster form:

(i) Given that:

Hence,
(ii) Given that

Hence,
(iii) Given that
So the positive factors of prime number are 1 and . Hence

Question:

Write the following sets in the roaster form:

(i) Given that $D = \left\{t : t^3 = t, t\in R \right \}$

hence,
(ii) Given that

Hence,
(iii) Given that

and
Hence,

Question:

If . Write Y in the roaster form.

Given that

The factors of are
The factors of are 1 and
Hence,

Question:

State which of the following statements are true and which are false. Justify your answer.
(i)
(ii)
(iii)
(iv)

(i) Given that:
Factors of 35 are 1, 5, 7, 35.
Hence, true
(ii) Given that:
Factors of 128 are 1,2,4,8,16,32,64,128
sum of all factors
hence statement is false.
(iii) Given that:
$x^4 - 5x^3 + 2x^2 -112x + 6 = 0$
Put x = 3
which is not true, So 3 is not an element of the set
Hence, statement is true
(iv) Given that:
The positive factors of 496 are 1,2,4,8,16, 31, 62, 124, 248, and 496
The sum of all positive factors is which is true, so the 496 is element of the set . Hence the statement is false.

Question:

Given and . Verify that .

Given that and .
Now,

\begin{aligned} RHS& = (L-M)\cap (L-N) \\ (L-M) &= \{1,2,3,4\} - \{3,4,5,6\} = \{1,2\} \\ (L-N) &= \{1,2,3,4\} - \{1,3,5\} = \{2,4\} \\ (L-M) & \cap (L-N) = \{1,2\} \cap \{2,4\} = \{2\} \\ LHS & = RHS\end{aligned}

Question:

If A and B are subsets of the universal set U, then show that
(i) A ⊂ A ∪ B
(ii) A ⊂ B ⇔ A ∪ B = B
(iii) (A ∩ B) ⊂ A

(i) Given that and
To prove: we have to show that if
Let
hence,
(ii) then let

But
From (i) and (ii) we get
Now if , then
let
Hence,
Thus,
iii)
$\begin{array}{l} \text { Let } x \in A \cap B \\ \quad x \in A \text { and } x \in B \\ \quad x \in A \\ \text { Hence, } A \cap B \subset A \end{array}$

Question:

Given that N = {1, 2, 3,…, 100}. Then write
(i) the subset of N whose elements are even numbers.
(ii) the subset of N whose element are perfect square numbers.

N = {1,2,3,4……100}

1. Required subset whose elements are even = {2,4, 6,8…..,100}

2. Required subset whose elements are perfect squares = {1,4,9 ,16,25,36,49,64,81,100}

Question:

A, B and C are subsets of Universal Set U. If A = {2, 4, 6, 8, 12, 20}, B = {3, 6, 9, 12, 15}, C = {5, 10, 15, 20} and U is the set of all whole numbers, draw a Venn diagram showing the relation of U, A, B and C.

Given that: A, B and C are the subsets of a universal set U.
Where A= {2,4,6,8,12,20}
B = {3, 6,9, 12, 15}
And C = {5, 10, 15, 20}

Question:

For all sets A, B and C, show that

Let

So,
Let

So,
From (i) and (ii), we get,

Question:

For all sets A and B,

\begin{aligned}LHS &= (A-B)\cup (A\cap B) \\ &= [(A-B)\cup A]\cap [(A-B)\cup B] \\ &=A\cap (A\cup B) = A = RHS \end{aligned}

Question:

For all sets A, B and C, A – (B – C) = (A – B) – C

Let us use the following Venn diagram to solve this question
Step 1: - B - C

Step 2: - A-(B-C)

Step 3: - A-B

Step 4: - (A-B)-C

As steps 2 and 4 are inequal, hence, the statement given is not true.

Question:

For all sets A, B and C, if A ⊂ B, then A ∩ C ⊂ B ∩ C

Suppose ,
Let

Hence the given statement is true.

Question:

For all sets A, B and C, if A ⊂ B, then A ∪ C ⊂ B ∪ C

Suppose ,
Let

Hence the given statement is true.

Question:

For all sets A, B and C, if A ⊂ C and B ⊂ C, then A ∪ B ⊂ C

Suppose .
Let

hence the given statement is true

Question:

For all sets A and B, A ∪ (B – A) = A ∪ B

Hence the given statement is true.

Question:

For all sets A and B, A – (A – B) = A ∩ B

]
Hence, the given statement is proved.

Question:

Let

Is T an empty set? Justify your answer.

Given:
\begin{aligned}\Rightarrow& & \frac{x+5}{x-7}- 5 &= \frac{4x-40}{13-x} \\ && \frac{(x+5)-5(x-7)}{x-7} & = \frac{4x-40}{13-x} \\ \Rightarrow && \frac{x + 5 - 5x + 35}{x-7} & = \frac{4x-40}{13-x} \\ \Rightarrow & & -4(x-10)(13-x) &= 4( x -10)(x-7) \\ \Rightarrow & & -4(x-10) (13-x + x -7) & = 0 \\ \Rightarrow & & -4(x - 10)6 &= 0 \\ \Rightarrow & & -24(x-10) &=0 \\ \Rightarrow & & x - 10 &= 0 \\ \Rightarrow & & x = 10& \text{ or }T = \{10\}\end{aligned}
Hence T is not an empty set.

## NCERT Exemplar Class 11 Maths Solutions Chapter 1: Long Answer Type

Question:

Let A, B and C be sets. Then show that A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)

Let
\begin{aligned} \Rightarrow & x\in A \text{ and } x\in (B\cup C) \\ \Rightarrow & x\in A \text{ and } x\in B\text{ or }x\in C \\ \Rightarrow & x\in A \text{ and } x\in B\text{ or }x\in A \text{ and } x\in C \\ \Rightarrow & x\in(A\cap B)\text{ or } x\in (A\cap C) \\ \Rightarrow & x\in (A\cap B)\cup (A\cap C) \\ \therefore & A\cap (B\cup C) \subset \ (A\cap B)\cup (A\cap C) & -(i) \end{aligned}
Let
\begin{aligned} \Rightarrow & y\in (A\cap B)\cup(A\cap C) \\ \Rightarrow & y \in (A\cap B)\text{ or }y\in (A\cap C) \\ \Rightarrow & y\in A\text{ and } y\in B\text{ or }y\in A\text{ and } y\in C\\ \Rightarrow &y \in A\text{ and }(y \in B\text{ or }y\in C) \\ \Rightarrow &y \in A\text{ and }(y \in B\cup C )\\ \Rightarrow &y \in A\cap(B\cup C ) \\\therefore & \ (A\cap B)\cup (A\cap C)\subset A\cap (B\cup C) & -(ii)\end{aligned}
From and (i) and (ii)

Question:

Out of 100 students; 15 passed in English, 12 passed in Mathematics, 8 in Science, 6 in English and Mathematics, 7 in Mathematics and Science; 4 in English and Science; 4 in all the three. Find how many passed
(i) in English and Mathematics but not in Science
(ii) in Mathematics and Science but not in English
(iii) in Mathematics only
(iv) in more than one subject only

Let the set of students who passed in mathematics be M, the set of students who passed in English be E and set of students who passed in science be S.

Then n(U)=100, n(M)=12, n(S)=8,
and
According to the Venn diagram

$n(E\cap M)=6; d + e= 6; d = 2$

$n(E\cap S)=4; b + e= 4; b= 0$

$n(M)=12; d + e + f+g= 12; g= 3$
$n(S)=8; b + e + f + c= 8; c= 1$
Thus, number of students who passed in English and mathematics but not in science = d= 2
Number of students who passed in Science and mathematics but not in English = f=3
Number of students who passed in mathematics only = g =3
Number of students who passed in more than one subject = b+ e+ d +f =0+4+2+3 =9

Question:

In a class of 60 students, 25 students play cricket and 20 students play tennis, and 10 students play both the games. Find the number of students who play neither?

Let C be the set of students who play cricket and T the set of students who play tennis.
Total number of students = 60 ⇒n(U)=60
Number of students who play cricket = 25 n(C)=25
Number of students who play tennis = 20 n(T)=20
Number of students who play both the games = 10
Number of students who play any one game
Number of students who play neither

Question:

In a town of 10,000 families it was found that 40% families buy newspaper A, 20% families buy newspaper B, 10% families buy newspaper C, 5% families buy A and B, 3% buy B and C and 4% buy A and C. If 2% families buy all the three newspapers. Find
(a) The number of families which buy newspaper A only.
(b) The number of families which buy none of A, B and C

Let A be a set of families which buy newspaper A, B be the set of families which buy newspaper B and C be the set of families which buy newspaper C
$n(U) = 10000, n(A) = 40\%, n(B) = 20\%,n(c) = 10\%, n(A\cap B) = 5\%, n(B\cap C) = 3\%,$ and
Number of families which buy newspaper A only
$= 10000 \times (\frac{40}{100} - \frac{5}{100} - \frac{4}{100} +\frac{2}{100}) = 10000\times \frac{33}{100} = 3300 \ \text{families}$
Number of families which buy none of A, B and C

$\small = [100 - (40 + 20 + 10 - 5-3-4+2)]\times 10000$
$\small = \frac{40}{100}\times 10000 = 4000 \ families$

Question:

In a survey of 200 students of a school, it was found that 120 study Mathematics, 90 study Physics and 70 study Chemistry, 40 study Mathematics and Physics, 30 study Physics and Chemistry, 50 study Chemistry and Mathematics and 20 none of these subjects. Find the number of students who study all the three subjects.

Let the set of students who study mathematics be M, the set of students who study physics be P and set of students who study chemistry be C.
N(U) =200, n(M)=120, n(P)=90, n(C)=70, n(MP) = 40, , and

$n(M \cup P \cup C) = n(M) + n(P) + n(C) - n(M\cap P) - n(P\cap C)$

Hence, the number of students who study all 3 subjects = 20

Question:

In a group of 50 students, the number of students studying French, English, Sanskrit were found to be as follows:
French = 17, English = 13, Sanskrit = 15
French and English = 09, English and Sanskrit = 4
French and Sanskrit = 5, English, French and Sanskrit = 3. Find the number of students who study
(i) French only
(ii) English only
(iii) Sanskrit only
(iv) English and Sanskrit but not French
(v) French and Sanskrit but not English
(vi) French and English but not Sanskrit
(vii) at least one of the three languages
(viii) none of the three languages

Let F be the set of students who study French, E be the set of students who study English and S be the set of students who study Sanskrit
and

Number of students who study French only =a =6
Number of students who study English only =c =3
Number of students who study Sanskrit only =g =9
Number of students who study English and Sanskrit but not French =f =1
Number of students who study French and Sanskrit but not English =d =2
Number of students who study French and English but not Sanskrit =b = 6
Number of students who study at least one of the three languages = a+b+c+d+e+f+g =6+6+3+2+3+1+9 =30
Number of students who study none of the three languages = 50-30=20

## NCERT Exemplar Class 11 Maths Solutions Chapter 1: Objective Type

Question:

Suppose are thirty sets each having 5 elements and are n sets each with 3 elements, let

and each element of S belongs to exactly 10 of the Ai’s and exactly 9 of the B,’S. then n is equal to
A. 15
B. 3
C. 45
D. 35

Number of elements in

However, each element is repeated 10 times.

Number of elements in
But each element is repeated 09 times

From (i) and (ii) we get . hence, the correct option is (c)

Question:

According to the question we have,
$\\2^m -2^n = 112 \\ \quad 2^n(2^{m-n} - 1) = 2^4\times 7\\ 2^n = 2^4 \text{ and } 2^{m-n} -1 = 7\\ n = 4\text{ and } 2^{m-n} = 1+7 = 8 = 2^3 \\ n+4\text{ and } m-n = 3 \\ m = 7$

Question:

Let R be set of points inside a rectangle of sides a and b (a, b > 1) with two sides along the positive direction of x-axis and y-axis. Then
A.
B.
C.
D.

The answer is the option (d)
R be set of points inside a rectangle of sides a and b (a, b > 1) with two sides along the positive direction of x-axis and y-axis
$R = \left\{(x,y) : 0 < x< a, 0< y< b\right\}$
Hence, the correct option is (d)

Question:

In a class of 60 students, 25 students play cricket and 20 students play tennis, and 10 students play both the games. Then, the number of students who play neither is
(a) 0 (b) 25 (c) 35 (d) 45

The answer is the option (b).
Let C be the set of students who play cricket and T the set of students who play tennis.
Total number of students = 60 ⇒n(U)=60
Number of students who play cricket = 25 n(C)=25
Number of students who play tennis = 20 n(T)=20
Number of students who play both the games = 10
Number of students who play any one game
Number of students who play neither

Question:

In a town of 840 persons, 450 persons read Hindi, 300 read English and 200 read both. Then the number of persons who read neither is
(a) 210 (b) 290 (c) 180 (d) 260

The answer is the option (b)
Total number of persons in a town = 840 n(U) = 840
Number of students who read Hindi = 450 n(H) = 450
Number of students who play English = 300 n(E) = 300
Number of students who read both = 200
\begin{aligned} n(H\cup E) = n(H) + n(E) - n(H\cap E) &= 450+300 - 200 = 550 \\ n(H'\cap E') = n(U) - n(H\cup E) &= 840 - 550 = 290\end{aligned}

Question:

If } and . Then
(A) (B) (C) X = Y (D)

The answer is the option (a).
Given that
And
Here, it is clear that every element belonging to the X is also present in Y.
. Hence, the correct option is (A).

Question:

A survey shows that 63% of the people watch a News Channel whereas 76% watch another channel. If x% of the people watch both channel, then
(A) x = 35 (B) x = 63 (C) (D) x = 39

Let p% of people watch a channel and q% of people watch another channel

So,

$x \geq 39$
Now n(p) = 63.

So,
Hence, the correct option is (C).

Question:

If sets A and B are defined as , then
(A) A ∩ B = A (B) A ∩ B = B (C) A ∩ B = φ (D) A ∪ B = A

The answer is option (c). Given that and
So, and

Thus, option C is correct.

Question:

If A and B are two sets, then A ∩ (A ∪ B) equals
(A) A (B) B (C) (D) (A ∪ B)

The answer is the option (a). Let

Hence A is correct.

Question:

If A = {1, 3, 5, 7, 9, 11, 13, 15, 17} B = {2, 4, ... , 18} and N the set of natural numbers is the universal set, then A′ ∪ (A ∪ B) ∩ B′) is
(A) φ (B) N (C) A (D) B

The answer is the option (b).
Given that A ={1,3,5,7,9,11,13,15,17}
B = {2,4,6………..,18}
U=N={1,2,3,4,5,……..}
\begin{aligned}A'\cup (A\cup B)\cap B' &= A'\cup [(A\cap B')\cup (B\cap B')] \\ & = A'\cup (A\cap B')\cup \phi&[A\cap A' = \phi] \\ &= A'\cup(A\cap B') \\ & = (A'\cup A) \cap (A'\cup B') \\ &=N\cup (A'\cup B')\\& = A'\cup B' \\ & = (A\cap B)' = \phi ' = N\end{aligned}

Question:

Let S = {x | x is a positive multiple of 3 less than 100} P = {x | x is a prime number less than 20}. Then n(S) + n(P) is
(A) 34 (B) 41 (C) 33 (D) 30

The answer is the option (b).
Given that: S={x| x is a positive multiple of 3 < 100}
S = {3, 6, 9, 12, 15, 18 …..99}
n(S) = 33
T = { x| x is a prime number < 20}
T = {2, 3, 5, 7, 11, 13, 17, 19}
n(T) = 8
So, n(S) + n(T) = 33+8 =41
Hence, (b) is correct.

Question:

The set can be written as ______________.

The set can be written as [1,2)

Question:

When, then number of elements in P(A) is ______________.

When, then number of elements in P(A) is 1
Here , n(A) = 0

Question:

If A and B are any two sets, then A – B is equal to

From the Venn diagram

Question:

Given the sets A = {1, 3, 5}. B = {2, 4, 6} and C = {0, 2, 4, 6, 8}. Then the universal set of all the three sets A, B and C can be ______________.

Given that A = {1,3,5}, B= {2,4,6} and C = {0,2,4,6,8}
Universal set of all given sets is U = {0, 1, 2, 3, 4, 5, 6, 8}

Question:

If U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}, A = {1, 2, 3, 5}, B = {2, 4, 6, 7} and C = {2, 3, 4, 8}. Then
(i) (B ∪ C)′ is ______________. (ii)(C – A)′ is ______________.

Given that: U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}, A = {1, 2, 3, 5}, B ={2, 4, 6, 7} and C = {2, 3, 4, 8}

Question:

If A is any set, then A ⊂ A

Every set is a subset of itself. Hence, true.

Question:

Given that M = {1, 2, 3, 4, 5, 6, 7, 8, 9} and if B = {1, 2, 3, 4, 5, 6, 7, 8, 9}, then

M = {1, 2, 3, 4, 5, 6, 7, 8, 9}
And B = {1, 2, 3, 4, 5, 6, 7, 8, 9}
Since B and M has same elements
M = B, So, is false

Question:

The sets {1, 2, 3, 4} and {3, 4, 5, 6} are equal.

The two sets do not contain same elements. Hence, false.

Question:

, where Q is the set of rational numbers and Z is the set of integers.

Since every integer is a rational number

Hence true

Question:

Let sets R and T be defined as

Then T ⊂ R

R = {………….., -8, -6, -4, -2, 0, 2, 4, 6, 8,…..}
And T= {……., -18, -12, -6, 0 , 6, 12, 18,…….}
Since every element of T is present in R. So,
Hence, the statement is true.

Question:

Here A = { 0, 1, 2} B is a set having all real numbers from 0 to 2.
So, $A\neq B$
Hence, the given statement is False.

Question:

Match the following sets for all sets A, B and C

1. (a)

2. (b)

3. (c)

4. (d)

5. (e)

6. (f)

(I)

(II)
\begin{aligned} {[B'\cup (B'-A')]'} &=[B'\cup (B'\cap A')]' \\ &=(B')'\cap (B'\cap A')'\\ & = B\cap (B\cup A) \\ & = B \end{aligned}

(iii)

(iv)

(v)
(vi)
So (i) – b , (ii) – c, (iii) – a, (iv) – (f), (v) - (d), (vi) – (e)

## Important Notes To Follow While Studying NCERT Exemplar Class 11 Maths Solutions Chapter 1 Sets

Sets are basically defined as a collection of objects belonging to a particular field or are of the same kind in a defined manner and could be differentiated from others for all at the same time or in the same sense.

For example- a group of days in a week will be the same for everyone, so, it is considered a set whereas a group of top ten favourite musicians in the world would be different for everyone as everyone has different choices, so this collection or group is not a set.

NCERT Exemplar Class 11 Maths chapter 1 solutions provided here for the NCERT books are very useful for Board exams as well as the JEE Main exams. Students can use NCERT Exemplar Class 11 Maths solutions chapter 1 PDF download for offline learning.

## Main Subtopics of NCERT Exemplar Class 11 Maths Solutions Chapter 1

The topics included in the chapter are mentioned below:

• 1.1 Introduction
• 1.2 Sets and their Representations
• 1.3 The Empty Set
• 1.4 Finite and Infinite Sets
• 1.5 Equal Sets
• 1.6 Subsets
• 1.6.1 Subsets of a set of real numbers
• 1.6.2 Intervals as subsets of R
• 1.7 Power Set
• 1.8 Universal Set
• 1.9 Venn Diagrams
• 1.10 Operations on Sets
• 1.10.1 Union of sets
• 1.10.2 Intersection of sets
• 1.10.3 Difference of sets
• 1.11 Complement of a Set
• 1.12 Practical Problems on Union and Intersection of Two Sets

## What will the students learn in NCERT Exemplar Class 11 Maths Solutions Chapter 1?

With NCERT Exemplar Class 11 Maths chapter 1 solutions, the students will learn about forming sets through a roaster and set builder method. This will help in various life situations to form combinations or groups of things belonging to the same category or are related to one another with a well-defined purpose. They also have wide application in mathematics for different subjects and are used in every field of mathematics in some or the other way. It also has its application in different programming languages and economic analysis, which will enable students to relate its use to real life. NCERT Exemplar Class 11 Maths solutions chapter 1 will also teach discipline to students through organizing and categorizing different things for better daily needs.

## NCERT Solutions for Class 11 Mathematics Chapters

 Chapter 1 Sets Chapter 2 Relations and Functions Chapter 3 Trigonometric Functions Chapter 4 Principle of Mathematical Induction Chapter 5 Complex Numbers and Quadratic Equations Chapter 6 Linear Inequalities Chapter 7 Permutations and Combinations Chapter 8 Binomial Theorem Chapter 9 Sequences and Series Chapter 10 Straight lines Chapter 11 Conic Sections Chapter 12 Introduction to Three Dimensional Geometry Chapter 13 Limits and Derivatives Chapter 14 Mathematical Reasoning Chapter 15 Statistics Chapter 16 Probability

## Important Topics To Cover in NCERT Exemplar Class 11 Maths Solutions Chapter 1 Sets

· Class 11 Maths NCERT Exemplar solutions Chapter 1 also provides two different methods for the representation of sets, i.e. roster or tabular form and set builder form both of which use different representations for displaying a set. The roster of the tabular method shows elements in a form, as A = {a, e, w, o, u}, but the same set in set-builder form will be represented as A = {a : a is a vowel in English language} which depicts that the element an of set A is a vowel of English language.

· NCERT Exemplar solutions for class 11 Maths chapter 1 also defines different types of the set such as empty set not having any member elements, finite set, having fixed/countable elements, infinite set, having uncountable member elements, equal sets have identical elements, power set, universal set, and subsets.

· NCERT Exemplar Class 11 Maths solutions chapter 1 also covers Venn diagrams and its application in different operations of the set, including union, intersection, and difference along with their detailed properties.

Check Chapter-Wise NCERT Solutions of Book

 Chapter-1 Sets Chapter-2 Relations and Functions Chapter-3 Trigonometric Functions Chapter-4 Principle of Mathematical Induction Chapter-5 Complex Numbers and Quadratic equations Chapter-6 Linear Inequalities Chapter-7 Permutation and Combinations Chapter-8 Binomial Theorem Chapter-9 Sequences and Series Chapter-10 Straight Lines Chapter-11 Conic Section Chapter-12 Introduction to Three Dimensional Geometry Chapter-13 Limits and Derivatives Chapter-14 Mathematical Reasoning Chapter-15 Statistics Chapter-16 Probability

### NCERT Exemplar Class 11 Solutions

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### Frequently Asked Question (FAQs)

1. Which topics are covered in this chapter?

The chapter of sets covers everything related to sets like the types, the subsets, the Venn diagrams and practical problems related to sets.

2. Can these solutions be helpful for competitive exams?

Yes, these questions and NCERT Exemplar Class 11 Maths solutions chapter 1 are helpful for those who are preparing for engineering and other entrance exams.

3. How to make the most of these solutions?

These NCERT exemplar Class 11 Maths chapter 1 solutions can be used while practising the questions from the NCERT book. These solutions can also be a reference for checking the answers.

4. How many questions are solved in these solutions?

Our team has solved every exercise question that is present in the main exercise and also the ones in the additional questions.

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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) Option 2) Option 3) Option 4)

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

 Option 1) Option 2) Option 3) Option 4)

In the reaction,

 Option 1)   at STP  is produced for every mole   consumed Option 2)   is consumed for ever      produced Option 3) is produced regardless of temperature and pressure for every mole Al that reacts Option 4) at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, 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