NCERT Exemplar Class 11 Maths Solutions Chapter 8 Binomial Theorem

# NCERT Exemplar Class 11 Maths Solutions Chapter 8 Binomial Theorem

Edited By Ravindra Pindel | Updated on Sep 12, 2022 04:16 PM IST

NCERT Exemplar Class 11 Maths Solutions Chapter 8 - Squares and roots solving have always been a major issue for a lot of students when solving algebra. For those who cannot use calculators in school, French mathematician Blaise Pascal removed the Binomial Theorem to make the process easier and faster. In NCERT exemplar Class 11 Maths solutions chapter 8, we discover a term known as the Pascal Triangle, which deals with how array arrangements of coefficients of expressions are for the derivation of the root answer. Integers and indices play an important role in this NCERT exemplar Class 11 Maths solutions chapter 8 where they are kept in formulas to find out answers. The positioning of the terms in the binomial theorem has a value and its importance, which is explained well in the General and Middle terms section. NCERT exemplar Class 11 Maths solutions chapter 8 pdf download can be availed using ‘webpage download as PDF online tools.
Also, check - NCERT Class 11 Maths Solutions for other chapters

JEE Main Scholarship Test Kit (Class 11): Narayana | Physics WallahAakash Unacademy

## NCERT Exemplar Class 11 Maths Solutions Chapter 8: Excercise-1.3

Question:1

Find the term independent of x, where x≠0, in the expansion of

............... (Given)
……. (from standard formula of Tr+1)
…… (i)
Now, for x,
30 – 3r = 0
→ r = 10
Substituting the value of r in eq (i),

Question:2

if the term free from x is the expansion of is 405, then find the value of k.

................... (Given)
……. (from standard formula of Tr+1)

…….. (i)
Now, for x,
(10 – 5r)/2 = 0
→ r = 2
Substituting the value of r in eq (i),

= 405

Thus k2 = 9
Thus, k = ±3

Question:3

Find the coefficient of x in the expansion of $\left ( 1-3x+7x^{2} \right )\left ( 1-x \right )^{16}$

(1-3x+7x2) (1-x)16 ….. (given)
= (1-3x+7x2) (106C0 - 16C1x1 + 16C2x2 + …… + 16C16x16)
= (1-3x +7x2) (1-16x+120x2+…..)
Thus, coefficient of x = -19

Question:4

Find the term independent of x in the expansion of

Given: (3x – 2/x2)15
Thus, $T_{r+1} = ^{15}C_r3^{15-r}x^{15-3r}(-2)^r$
Now, for independent term x,
15 – 3r = 0
Thus, r = 5
Thus, the term x is
T5+1 = 15C5315-5(-2)5
= -15 x 14 x 13 x 12 x 11 x 10!/5x4x3x2x1x10! .310.25
= -3003 x 310x 25

Question:5

Find the middle term (terms) in the expansion of
i) ii)

(i) Given: (x/a – a/x)10
Where, n = 10(even).
The middle term = (10/2 + 1)th term, viz., the sixth term
Thus, T6 = T5+1 10C5 (x/a)10-5 (-a/x)5
= - 10C5 (x/a)5(a/x)5
= - 10 x 9 x 8 x 7 x 6 x 5!/ 5! x 4 x 3 x 2 x 1 (x/a)5(x/a)-5
= -252
(ii) Given: (3x – x3/6)9
Where, n = 9 (odd)
Now, here are two middle terms-
(9+1/2)th ,viz., the tenth term
& (9+1/2 + 1)th, viz., the sixth term
Thus, T5 = T4+1
= 9C4 (3x)9-4 (-x3/6)4
= 9x8x7x6x5!/4x3x2x1x5! . 35x5x126-4
= 189/8 x17
Now,
T6 = T5+1
= 9C5(3x)9-5 (-x3/6)5
= - 9 x 8 x 7 x 6 x 5!/ 5! x 4 x 3 x 2 x 1 . 34.x4.x15.6-4
= - 21/16 x19

Question:6

Find the coefficient of x15 in the expansion of (x – x2)10

Given: (x – x2)10
Thus, Tr+1 = 10Cr x10-r(-x2)r
= (-1)r 10Crx10-rx2r
= (-1)r 10Crx10+r
Now, for coefficient of x15
We have,
10 + r = 15
Thus, r = 5
Thus, T5+1 = (-1)5.10C5x15
Therefore,
Coefficient of x15 = -10x9x8x7x6x5!/5x4x3x2x1x5!
= -252

Question:7

Find the coefficient of x1/17 in the expansion of (x4 – 1/x3)15

Given: (x4 – 1/x3)15
Thus,
Tr+1 = 15Cr(x4)15-r(-1/x3)r
= 15Cr x60-4r(-1)r x-3r
= 15Cr x60-7r(-1)r
Now, for coefficient x-17,
60 – 7r = -17
7r = 77
r = 11
Thus, T11+1 = 15C11 x60-77(-1)11
Therefore,
Coefficient of x-17 = -15 x 14 x 13 x 12 x 11!/11! x 4 x 3 x 2 x 1
= -1365

Question:8

Given: (y1/2 + x1/3)n
From the end, binomial coefficient of 3rd term = 45
Thus, nCn-2 = 45 , nC2 = 45
i.e., n(n-1)(n-2)!/2!(n-2)! = 45
Now, n(n-1) = 90,
n2 – n – 90 = 0,
(n – 10)(n + 9) = 0
n = 10 …. (since n is not equal to -9)
Now, the 6th term,
=10C5 (y1/2)10-5 (x1/3)5
= 252y5/2.x5/3

Question:9

Find the value of r, if the coefficients of (2r + 4)th and (r – 2)th terms in the expansion of (1 + x)18 are equal.

Given: (1+x)18
Now, T(2r+3)+1 is the (2r + 4)th term
Thus, T(2r+3)+1 = 18C2r+3 (x)2r+3
Now, T(r-3)+1 is the (r-2)th term
Thus, T(r-3)+1 = 18Cr-3 xr-3
Now, it is given that the terms are equal in expansion,
Thus,
18C2r+3 = 18Cr-3
i.e., 2r + 3 + r – 3 = 18 …… (nCx = nCy gives x + y = n)
Thus, 3r = 18
Therefore, r = 6

Question:10

Given: (1+x)2n
Now, 2nC1, 2nC2, 2nC3 are the coefficients of the first second and the third terms respectively.
It is also given that the coefficients are in A.P.
Thus, 2. 2nC2 =2nC1 + 2nC3
Thus, 2[2n(2n-1)(2n-2)!/2x1x(2n-2)!] = 2n + 1n(2n-1)(2n-2)(2n-3)!/3!(2n-3)!
N(2n-1) = n = n(2n-1)(n-1)/3
3(2n-2) = 3+ (2n2 – 3n + 1)
6n – 3 = 2n2 – 3n + 4
Thus, 2n2 – 9n + 7 = 0

Question:11

Find the coefficient of x4 in the expansion of (1 + x + x2 + x3)11.

Given: (1+x+x2+x3)11
That is equal to [(1+x) + x2(1 + x)]11
= [(1+x) (1+x2)]11
= (1+x)11.(1+x2)11
= (11C0 + 11C1x + 11C2x2 + 11C3x3 + 11C4x4 + ….) (11C0 + 11C1x2 + 11C2x4 + ….)
Thus, the coefficient of x4 =11C0 x 11C4 + 11C1 x 11C2+11C2x 11C0
= 330 + 605 + 55
= 990

Question:12

If p is a real number and the middle term in the expansion (p/2 + 2)8 is 1120, then find the value of p.

Given: (p/2 + 2)8
Now, here index is n = 8,
Since there is only one middle term, viz., (8/2 + 1)th i.e., the fifth term
T5 = T4+1
= 8C4(p/2)8-4. 24
Now,
1120 = 8C4p4
1120 = 8x7x6x5x4!/4!x4x3x2x1 . p4
1120 = 7x2x5xp4
Thus, p4 = 1120/70
p4= 16
p2 = 4
Thus, p = ±2

Question:13

Given: (x – 1/x)2n
Now, here the index = 2n(even)
There’s only one middle term, i.e., (2n/2 + 1)th term, viz., (n + 1)th term
Tn+1 = 2nCn(x)2n-n(-1/x)n
= 2nCn(-1)n
= (-1)n(2n!)/n!.n!
= (-1)n 1.2.3.4.5. …. (2n-1).(2n)/n!.n!
= (-1)n [1.3.5…. (2n-1)].2n[1.2.3…..n]/(1.2.3…n).n!
= (-2)n[1.3.5…(2n-1)].2n/n!

Question:14

Find n in the binomial if the ratio of 7th term from the beginning to the 7th term from the end is

Given
Now, from the beginning,
Seventh term is
...................(i)
& from the end,
The seventh term is the same ,i.e.,
.............(ii)
Now, it is given that

Thus

Thus

Thus $6^\frac{n-12}{3}=6^{-1}$

Thus n=9

Question:15

In the expansion of (x + a)n if the sum of odd terms is denoted by O and the sum of even term by E.
Then prove that
(i) O2– E2= (x2– a2)n (ii) 4OE = (x + a)2n- (x - a)2n

(i) (x+a)n= nC0xn + nC1xn-1 a1 + nC2xn-2 a2 + nC3xn-3 a3 + ….. + nCn an
Now, sum of odd terms,
O = nC0xn+ nC2xn-2 a2 + …….
& sum of even terms,
E = nC1xn-1 a1 + nC3xn-3 a3 + …..
Now, (x+a)n = O+E & (x-a)n = O – E
Thus, (O + E)(O – E) = (x + a)n(x – a)n
Thus, O2 – E2 = (x2 – a2)n
(ii)Here,
4OE = (O + E)2 – (O – E)2
= [(x+a)n]2 – [(x-a)n]2
= (x+a)2n – (x-a)2n

Question:16

If xn occurs expantion of then prove that its coefficient is

Given: (x2 + 1/x)2n
Thus, Tr+1 =2nCr(x2)2n-r(1/x)r
= 2nCrx4n-3r
Now, if xn will occur in the expansion, then,
Let us consider 4n – 3r = p
Thus,
3r = 4n – p
r = 4n – p/3
Thus, coefficient of xp = 2nCr
= (2n)!/r!(2n-r)!
=

Question:17

Find the term independent of x in the expansion of(1+x+2x3)(3x2/2 – 1/3x)9

Given: (1+x+2x3)( 3x2/2 – 1/3x)9
Now, let us consider (3x2/2 – 1/3x)9
Tr+1 = 9Cr (3/2 x2)9-r (-1/3x)r
= 9Cr (3/2)9-r (-1/3)r x18-3r
Thus, in the expansion of (1+x+2x3)(3/2x2 – 1/3x)9 the general term is:
9Cr (3/2)9-r (-1/3)r x18-3r + 9Cr (3/2)9-r (-1/3)r x19-3r + 2.9Cr (3/2)9-r (-1/3)r x21-3r
Now, for the independent term x,
Let us put 18 – 3r = 0,
19 – 3r = 0
& 21 – 3r = 0,
We get,
R = 6 & r = 7
Now, since the second term is not independent of x, it will be:
9C6 (3/2)9-6 (-1/3)6 + 2.9Cr (3/2)9-7 (-1/3)7
= 9x8x7x6!/6!x3x2 . 1/23.33 – 2. 9x8x7!/7!x2x1. 32/22.1/37
= 84/8.1/33 – 36/4.2/35
= 17/54

Question:18

The total number of terms in the expansion of (x + a)100 + (x – a)100 after simplification is
(A) 50 (B) 202 (C) 51 (D) none of these

(c) 51.
Now, (x+a)100 + (x-a)100 = (100C0x100 + 100C2x99a + 100C2x98a2 + ….) + (100C0x100 - 100C2x99a + 100C2x98a2)
= 2(100C0x100 + 100C2x98a2 +…..+100C100a100)
Therefore, there are 51 terms.

Question:19

Given the integers r > 1, n > 2, and coefficients of (3r)th and (r + 2)nd terms in the binomial expansion of (1 + x)2n are equal, then
(A) n = 2r (B) n = 3r (C) n = 2r + 1 (D) none of these

(a) n =2r
Given: (1+x)2n
Thus, T3r = T(3r-1)+1
= 2nC3r-1x3r-1
& Tr+2 = T(r+1)+1
= 2nCr+1xr+1
Now, 2nC3r-1x3r-1 = 2nCr+1xr+1 ……… (given)
Thus, 3r – 1 + r + 1 = 2n
Thus, n = 2r

Question:20

The two successive terms in the expansion of (1 + x)24 whose coefficients are in the ratio 1: 4 are
(A) 3rd and 4th (B) 4th and 5th (C) 5th and 6th (D) 6th and 7th

(c) 5th & 6th
Let us consider (r+1)th& (r+2)th as the 2 successive terms in the expansion of (1+x)24
Now, Tr+1 = 24Crxr& Tr+2 = 24Cr+1xr+1
Now, 24Cr/24Cr+1 = ¼ ……. (given)
Thus,

Thus, (r+1)r! (23 – r)! / r!(24 – r)(23 – r)! = ¼
r + 1/24 – r = ¼
4r + 4 = 24 – r
t = 4
Thus, T4+1 = T5
& T4+2 = T6
Thus, opt (c).

Question:21

The coefficients of xn in the expansion of (1 + x)2n and (1 + x)2n ~1 are in the ratio
(a) 1 : 2
(b) 1 : 3
(c) 3 : 1
(d) 2:1

The answer is the option (d) 2:1
2nCn is the coefficient of xn in the expansion of (1+x)2n
&2n-1Cn is the coefficient of xn in the expansion of (1+x)2n-1
$\frac{^{2n}C_n}{^{2n-1}C_n}=\frac{\frac{\left ( 2n \right )!}{\left ( n \right )!n!}}{\frac{\left ( 2n-1 \right )!}{n!\left ( n-1 \right )!}}$
= (2n)!n!(n-1)!/(2n-1)!n!n!
= 2n(2n-1)!n!(n-1)!/n!n(n-1)!(2n-1)!
= 2/1
= 2:1

Question:22

The answer is the option (b) 7
(1+x)n = nC0 + nC1x + nC2x2 + nC3x3 + ….. + nCnxn
Now, nC1, nC2, nC3 are the coefficients of the second, third and fourth terms respectively.
2 nC2 =nC1 + nC3 …… since they are in A.P.
2[n!/(n-2)!2!] = n + n!/ 3!(n-3)!
2[n(n-1)/2!] = n + n(n-1)(n-2)/3!
(n-1) = 1 + (n-1)(n-2)/6
6n – 6 = 6 + n2 – 3n + 2
n2 – 9n + 14 = 0
(n-7)(n-2) = 0
Thus, n = 2 or n = 7
Now, n = 2 is not possible,
Thus, n = 7

Question:23

The answer is the option (b) 2
2nCn is the coefficient of xn in the expansion of (1+x)2n
Thus, A = 2nCn
&2n-1Cn is the coefficient of xn in the expansion of (1+x)2n-1
Thus, B = 2n-1Cn
Thus, A/B = 2nCn/2n-1Cn
= 2/1
=2:1

Question:24

If the middle term of is equal to then the value of x is
(a) (b) (c) (d)

The answer is the option (c) nπ + (-1)nπ/6
Now, n = 10 (even)
Thus, there is only one middle term i.e., the 6th term
Thus, T6 = T5+1
= 10C5(1/x)10-5 (x sin x)5
63/8 = 10C5 sin5x …… (given)
63/8 = 252 x sin5x
Thus, sin5x = 1/32
Sin x = ½
Sin x = π/6
Thus, x = nπ + (-1)nπ/6

Question:25

30C30/2 = 30C15 is the largest coefficient in the expansion of (1+x)30

Question:27

In the expansion of (x2 – 1/x2)16, the value of constant term is_________________.

Here, Tr+1 = 16Cr(x2)16-r(-1/x2)r
= 16Crx32-4r(-1)r
Now, for constant term,
32 – 4r = 0
Thus, r = 8
Thus, T8+1 = 16C8

Question:28

If the seventh terms from the beginning and the end in the expansion of are equal, then n equals _________________

Given:
Now, referring to que 14.

Thus

Thus
6(n-12)/3 = 60
(n-12)/3 = 0
Thus, n = 12

Question:29

The coefficient of in the expansion of is ______________

Given: (1/a – 2b/3)10
Thus, Tr+1 = 10Cr(1/a)10-r (-2b/3)r
Now, for the coefficient of a-6b4,
10 – r = 6
Thus, r = 4
Therefore the coefficient of a-6b4 = 10C4(-2/3)4
= 10.9.8.7.6!/6!.4.3.2.1 . 24/34
= 1120/27

Question:30

Middle term in the expansion of (a3 + ba)28is _________

Given: (a3 + ba)28
Here, n = 28 (even)
Thus, there is only one middle term (28/2 + 1)th or 15th term.
Thus, the middle term, T15 = T14+1
= 28C14(a3)28-14(ba)14
= 28C14a42b14a14
= 28C14a56b14

Question:31

The ratio of the coefficients of xp and xq in the expansion of (1 +x)p+q is _______

Given: (1+x)p+q
Thus, coefficient of xp = p+qCp
& Coefficient of xq = p+qCq
Therefore,
p+qCp/ p+qCq = p+qCp /p+qCp
= 1:1

Question:32

The position of the term independent of x in the expansion of is _____________.

Given: $\left ( \sqrt{\frac{x}{3}}+\frac{3}{2x^{2}} \right )^{10}$
Thus, $T_{r+1}=^{10}C_{r}\left ( \sqrt{\frac{x}{3}} \right )^{10-r}\left ( \frac{3}{2x^{2}} \right )^{r}$
Now, for constant term,
10 – 5r = 0
Thus, r = 2
Therefore, the third term is independent of x.

Question:33

If 2515 is divided by 13, the reminder is _________ .

Now, 2515 = (26 – 1)15
= 15C02615 15C12614 + …. + 15C1426 – 15C15
= (15C0261515C12614 + …. + 15C1426 – 13) + 12
Therefore, it is clear that, when we divide 2515 by 13, we get 12 as the remainder.

Question:34

The sum of the series is

The given statement is False.
Now, from what’s given, we can say that,
= 20C0 + 20C1 + 20C2 + 20C3 + …… + 20C10
= 20C0 + 20C1 + ….. + 20C10 + 20C11 + …… + 20C20 – (20C11 + ….. + 20C20)
= 220 – (20C11 + …… + 20C20)

Question:35

The expression 79 + 97 is divisible by 64.

The given statement is True.
Given: 79 + 97 is divisible by 64.
Now,
79 + 97 = (1+8)7 – (1-8)9
= [7C0 + 7C1.8 + 7C2.(8)2 + 7C3.(8)3 + ….. + 7C7.(8)7] – [9C0 + 9C1.8 + 9C2.(8)2 - 9C3.(8)3 + ….. 9C9.(8)0]
= (7x8 + 9x8) + (21 x 82 – 36 x 82) + …..
= 128 + 64(21 – 36)
= 64[2 + (21 – 36) + …], viz. divisible by 64.

Question:36

The number of terms in the expansion of [(2x + y3)4]7 is 8.

The given statement is False.
Given: [(2x + y3)4]7 = (2x + y3)28
Thus, no. of terms = 28 + 1
= 29

Question:37

The sum of coefficients of the two middle terms in the expansion of (1+x)2n-1 is equal to 2n – 1Cn.

The given statement is False.
Given: (1 + x)2n-1
Now, no. of terms = 2n – 1 + 1
= 2n (even)
Thus, the middle term = 2n/2th terms & (2n/2 + 1)th terms
= nth terms & (n+1)th terms
Now, the coefficient of-
nth term = 2n – 1Cn-1
& (n + 1)th term = 2n-1Cn
Now, sum of the coefficients = 2n – 1Cn-1 + 2n-1Cn
= 2n-1Cn-1 + 2n-1Cn
= 2nCn

Question:38

The last two digits of the numbers 3400 are 01.

The given statement is true.
Given: 3400 = (9)200
= (10 – 1)200
Thus, (10 – 1)200 = 200C0(10)200200C1(10)199 + ….. – 200C199(10)1 + 200C200(1)200
= 10200 – 200 x 10199 + …. – 10 x 200 + 1
Therefore, the last two digits will be 01.

Question:39

If the expansion of (x-1/x2)2n contains a term independent of x, then n is a multiple of 2.

The given statement is False.
Given:
Now,
Tr+1 = 2nCr(x)2n-r (-1/x2)r
= 2nCr(x)2n-r-2r(-1)r
= 2nCr(x)2n-3r (-1)r(-1)r
Now, for the independent term of x,
2n – 3r = 0
Thus, r = 2n/3, viz. not an integer
Thus, this expression is not possible to be true.

Question:40

Number of terms in the expansion of (a + b)n where n ∈ N is one less than the power n.

The given statement is False.
In the given expression (a+b)n, the no. of terms is just 1 more than n, i.e., n + 1.

## NCERT Exemplar Class 11 Maths Solutions Chapter 8 Binomial Theorem- Topics

• 8.1 Introduction
• 8.2 Binomial Theorem for Positive Integral Indices
• 8.2.1 Binomial theorem for any positive integer n
• 8.2.2 Some special cases
• 8.3 General and Middle Terms

NCERT Exemplar Class 11 Maths Solutions Chapter-Wise

 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 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

## What will the students learn in NCERT Exemplar Class 11 Maths Solutions Chapter 8 Binomial Theorem?

Through NCERT Exemplar Class 11 Maths solutions chapter 8 will help students easily understand how to calculate roots and squares, with the help of the theorem. The slant array presentation will help them understand the pascal triangle with much ease and interest.

In NCERT exemplar solutions for class 11 Maths chapter 8, you will learn a very important statement about the theorem, i.e. that any positive integer can be termed as ‘n’, there will be a sum of any two numbers, of which n will be the power and how the sum of n+1 will derive. This is actually how the binomial theorem came into existence and is solved by people across. These solutions will ease the difficulty in finding root for students without the help of a calculator.

### NCERT Exemplar Class 11 Maths Solutions Chapter 8 - Important Topics

· Class 11 Maths NCERT Exemplar solutions chapter 8 has detailed that Squares and roots, binomial theorems, special cases and special terms are important topics which students should pay extra attention to.

· They will learn squares and roots with ease and score well in class. Major and complex problems can be easily solved with the help of this theorem. For students who want to pursue higher studies in this field should be very thorough while studying this chapter.

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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1. Are these solutions helpful in entrance exams?

Yes, these solutions help prepare the topics from base-up during entrance exam preparation.

2. How to use these solutions?

One can use these NCERT Exemplar Class 11 Maths chapter 8 solutions as a reference point while solving questions on one’s own.

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Yes, these NCERT Exemplar Class 11 Maths solutions chapter 8 are solved as per the NCERT pattern in the most detailed way as per the marking scheme.

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

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