NCERT Solutions for Exercise 5.1 Class 10 Maths Chapter 5 - Arithmetic Progressions

# NCERT Solutions for Exercise 5.1 Class 10 Maths Chapter 5 - Arithmetic Progressions

Edited By Ramraj Saini | Updated on Nov 17, 2023 04:51 PM IST | #CBSE Class 10th

## NCERT Solutions For Class 10 Maths Chapter 5 Exercise 5.1

NCERT Solutions for Exercise 5.1 Class 10 Maths Chapter 5 Arithmetic Progressions are discussed here. These NCERT solutions are created by subject matter expert at Careers360 considering the latest syllabus and pattern of CBSE 2023-24. This class 10 ex 5.1 the focus is on the general equation of arithmetic progression along with the general equation it includes many problems based on common difference and first term. Class 10 maths ex 5.1 also gives a very small idea about finite and infinite A.P series.

NCERT solutions for 10th class Maths exercise 5.1 answers are designed as per the students demand covering comprehensive, step by step solutions of every problem. Practice these questions and answers to command the concepts, boost confidence and in depth understanding of concepts. Students can find all exercise together using the link provided below.

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## Assess NCERT Solutions for Class 10 Maths Chapter 5 Exercise 5.1

Arithmetic Progressions Class 10 Chapter 5 Exercise: 5.1

(i) The taxi fare after each km when the fare is $\small Rs\hspace{1mm}15$ for the first km and $\small Rs\hspace{1mm}8$ for each additional $\small km$ .

It is given that
Fare for $1^{st} \ km$ = Rs. 15
And after that $\small Rs\hspace{1mm}8$ for each additional $\small km$
Now,
Fare for $2^{nd} \ km$ = Fare of first km + Additional fare for 1 km
= Rs. 15 + 8 = Rs 23

Fare for $3^{rd} \ km$ = Fare of first km + Fare of additional second km + Fare of additional third km
= Rs. 23 + 8= Rs 31

Fare of n km = $15 + 8 \times (n - 1)$
( We multiplied by n - 1 because the first km was fixed and for the rest, we are adding additional fare.

In this, each subsequent term is obtained by adding a fixed number (8) to the previous term.)

Now, we can clearly see that this is an A.P. with the first term (a) = 15 and common difference (d) = 8

(ii) The amount of air present in a cylinder when a vacuum pump removes $\small \frac{1}{4}$ of the air remaining in the cylinder at a time.

It is given that
vacum pump removes $\small \frac{1}{4}$ of the air remaining in the cylinder at a time
Let us take initial quantity of air = 1

Now, the quantity of air removed in first step = 1/4

Remaining quantity after 1 st step

$= 1-\frac{1}{4}= \frac{3}{4}$

Similarly, Quantity removed after 2 nd step = Quantity removed in first step $\times$ Remaining quantity after 1 st step

$=\frac{3}{4}\times \frac{1}{4}= \frac{3}{16}$
Now,

Remaining quantity after 2 nd step would be = Remaining quantity after 1 st step - Quantity removed after 2 nd step

$=\frac{3}{4}- \frac{3}{16}= \frac{12-3}{16}= \frac{9}{16}$
Now, we can clearly see that

After the second step the difference between second and first and first and initial step is not the same, hence

the common difference (d) is not the same after every step

Therefore, it is not an AP

(iii) The cost of digging a well after every meter of digging, when it costs $\small Rs\hspace{1mm}150$ for the first metre and rises by $\small Rs\hspace{1mm}50$ for each subsequent meter.

It is given that
Cost of digging of 1st meter = Rs 150
and
rises by $\small Rs\hspace{1mm}50$ for each subsequent meter
Therefore,

Cost of digging of first 2 meters = cost of digging of first meter + cost of digging additional meter

Cost of digging of first 2 meters = 150 + 50

= Rs 200

Similarly,
Cost of digging of first 3 meters = cost of digging of first 2 meters + cost of digging of additional meter

Cost of digging of first 3 meters = 200 + 50

= Rs 250

We can clearly see that 150, 200,250, ... is in AP with each subsequent term is obtained by adding a fixed number (50) to the previous term.

Therefore, it is an AP with first term (a) = 150 and common difference (d) = 50

(iv) The amount of money in the account every year, when $\small Rs\hspace{1mm}10000$ is deposited at compound interest at $\small 8\hspace{1mm}\%$ per annum .

Amount in the beginning = Rs. 10000

Interest at the end of 1st year at the rate of $\small 8\hspace{1mm}\%$
is $\small 8\hspace{1mm}\%$ of 10000 = $\frac{8\times 10000}{100}= 800$

Therefore, amount at the end of 1st year will be

= 10000 + 800

= 10800

Now,

Interest at the end of 2nd year at rate of $\small 8\hspace{1mm}\%$
is $\small 8\hspace{1mm}\%$ of 10800 = $\frac{8\times 10800}{100}= 864$

Therefore,, amount at the end of 2 nd year

= 10800 + 864 = 11664

Since each subsequent term is not obtained by adding a unique number to the previous term; hence, it is not an AP

$\small a=10,d=10$

It is given that
$\small a=10,d=10$
Now,
$a_1= a =10$
$a_2= a_1 + d= 10 + 10 = 20$
$a_3= a_2 + d= 20 + 10 = 30$
$a_4= a_3 + d= 30 + 10 = 40$
Therefore, the first four terms of the given series are 10,20,30,40

$\small a=-2,d=0$

It is given that
$\small a=-2,d=0$
Now,
$a_1= a = -2$
$a_2= a_1 + d= -2 + 0 = -2$
$a_3= a_2 + d= -2 + 0 = -2$
$a_4= a_3 + d= -2 + 0 = -2$
Therefore, the first four terms of the given series are -2,-2,-2,-2

$\small a=4,d=-3$

It is given that
$\small a=4,d=-3$
Now,
$a_1= a =4$
$a_2= a_1 + d= 4 - 3 = 1$
$a_3= a_2 + d= 1 - 3 = -2$
$a_4= a_3 + d= -2- 3 = -5$
Therefore, the first four terms of the given series are 4,1,-2,-5

$\small a=-1,d=\frac{1}{2}$

It is given that
$\small a=-1,d=\frac{1}{2}$
Now,
$a_1= a =-1$
$a_2= a_1 + d= -1 + \frac{1}{2} = -\frac{1}{2}$
$a_3= a_2 + d= -\frac{1}{2} + \frac{1}{2} = 0$
$a_4= a_3 + d= 0+\frac{1}{2}= \frac{1}{2}$
Therefore, the first four terms of the given series are $-1,-\frac{1}{2},0, \frac{1}{2}$

$\small a=-1.25,d=-0.25$

It is given that
$\small a=-1.25,d=-0.25$
Now,
$a_1= a =-1.25$
$a_2= a_1 + d= -1.25 -0.25= -1.50$
$a_3= a_2 + d= -1.50-0.25=-1.75$
$a_4= a_3 + d= -1.75-0.25=-2$
Therefore, the first four terms of the given series are -1.25,-1.50,-1.75,-2

$\small 3,1,-1,-3,...$

Given AP series is

$\small 3,1,-1,-3,...$

Now, first term of this AP series is 3

Therefore,

First-term of AP series (a) = 3

Now,

$a_1=3 \ \ and \ \ a_2 = 1$

And common difference (d) = $a_2-a_1 = 1-3 = -2$

Therefore, first term and common difference is 3 and -2 respectively

Given AP series is

$\small -5,-1,3,7,...$

Now, the first term of this AP series is -5

Therefore,

First-term of AP series (a) = -5

Now,

$a_1=-5 \ \ and \ \ a_2 = -1$

And common difference (d) = $a_2-a_1 = -1-(-5) = 4$

Therefore, the first term and the common difference is -5 and 4 respectively

$\small \frac{1}{3},\frac{5}{3},\frac{9}{3},\frac{13}{3},...$

Given AP series is

$\small \frac{1}{3},\frac{5}{3},\frac{9}{3},\frac{13}{3},...$
Now, the first term of this AP series is $\frac{1}{3}$

Therefore,

The first term of AP series (a) = $\frac{1}{3}$

Now,

$a_1=\frac{1}{3} \ \ and \ \ a_2 = \frac{5}{3}$

And common difference (d) = $a_2-a_1 = \frac{5}{3}-\frac{1}{3} = \frac{5-1}{3} =\frac{4}{3}$

Therefore, the first term and the common difference is $\frac{1}{3}$ and $\frac{4}{3}$ respectively

$\small 0.6,1.7,2.8,3.9,...$

Given AP series is

$\small 0.6,1.7,2.8,3.9,...$

Now, the first term of this AP series is 0.6

Therefore,

First-term of AP series (a) = 0.6

Now,

$a_1=0.6 \ \ and \ \ a_2 = 1.7$

And common difference (d) = $a_2-a_1 = 1.7-0.6 = 1.1$

Therefore, the first term and the common difference is 0.6 and 1.1 respectively.

$\small 2,4,8,12,...$

Given series is
$\small 2,4,8,12,...$
Now,
the first term to this series is = 2
Now,
$a_1 = 2 \ \ and \ \ a_2 = 4 \ \ and \ \ a_3 = 8$
$a_2-a_1 = 4-2 = 2$
$a_3-a_2 = 8-4 = 4$
We can clearly see that the difference between terms are not equal
Hence, given series is not an AP

$\small 2,\frac{5}{2},3,\frac{7}{2},...$

Given series is
$\small 2,\frac{5}{2},3,\frac{7}{2},...$
Now,
first term to this series is = 2
Now,
$a_1 = 2 \ \ and \ \ a_2 = \frac{5}{2} \ \ and \ \ a_3 = 3 \ \ and \ \ a_4 = \frac{7}{2}$
$a_2-a_1 = \frac{5}{2}-2 = \frac{5-4}{2}=\frac{1}{2}$
$a_3-a_2 = 3-\frac{5}{2} = \frac{6-5}{2} = \frac{1}{2}$
$a_4-a_3=\frac{7}{2}-3=\frac{7-6}{2} =\frac{1}{2}$
We can clearly see that the difference between terms are equal and equal to $\frac{1}{2}$
Hence, given series is in AP
Now, the next three terms are
$a_5=a_4+d = \frac{7}{2}+\frac{1}{2} = \frac{8}{2}=4$
$a_6=a_5+d = 4+\frac{1}{2} = \frac{8+1}{2}=\frac{9}{2}$
$a_7=a_6+d =\frac{9}{2} +\frac{1}{2} = \frac{10}{2}=5$

Therefore, next three terms of given series are $4,\frac{9}{2} ,5$

$\small -1.2,-3.2,-5.2,-7.2,...$

Given series is
$\small -1.2,-3.2,-5.2,-7.2,...$
Now,
the first term to this series is = -1.2
Now,
$a_1 = -1.2 \ \ and \ \ a_2 = -3.2 \ \ and \ \ a_3 = -5.2 \ \ and \ \ a_4 = -7.2$
$a_2-a_1 = -3.2-(-1.2) =-3.2+1.2=-2$
$a_3-a_2 = -5.2-(-3.2) =-5.2+3.2 = -2$
$a_4-a_3=-7.2-(-5.2)=-7.2+5.2=-2$
We can clearly see that the difference between terms are equal and equal to -2
Hence, given series is in AP
Now, the next three terms are
$a_5=a_4+d = -7.2-2 =-9.2$
$a_6=a_5+d = -9.2-2 =-11.2$
$a_7=a_6+d = -11.2-2 =-13.2$
Therefore, next three terms of given series are -9.2,-11.2,-13.2

$\small -10,-6,-2,2,...$

Given series is
$\small -10,-6,-2,2,...$
Now,
the first term to this series is = -10
Now,
$a_1 = -10 \ \ and \ \ a_2 = -6 \ \ and \ \ a_3 = -2 \ \ and \ \ a_4 = 2$
$a_2-a_1 = -6-(-10) =-6+10=4$
$a_3-a_2 = -2-(-6) =-2+6 = 4$
$a_4-a_3=2-(-2)=2+2=4$
We can clearly see that the difference between terms are equal and equal to 4
Hence, given series is in AP
Now, the next three terms are
$a_5=a_4+d = 2+4 =6$
$a_6=a_5+d = 6+4=10$
$a_7=a_6+d = 10+4=14$
Therefore, next three terms of given series are 6,10,14

$\small 3,3+\sqrt{2},3+2\sqrt{2},3+3\sqrt{2},...$

Given series is
$\small 3,3+\sqrt{2},3+2\sqrt{2},3+3\sqrt{2},...$
Now,
the first term to this series is = 3
Now,
$a_1 = 3 \ \ and \ \ a_2 = 3+\sqrt2 \ \ and \ \ a_3 = 3+2\sqrt2 \ \ and \ \ a_4 = 3+3\sqrt2$
$a_2-a_1 = 3+\sqrt2-3= \sqrt2$
$a_3-a_2 = 3+2\sqrt2-3-\sqrt2 = \sqrt2$
$a_4-a_3 = 3+3\sqrt2-3-2\sqrt2 = \sqrt2$
We can clearly see that the difference between terms are equal and equal to $\sqrt2$
Hence, given series is in AP
Now, the next three terms are
$a_5=a_4+d = 3+3\sqrt2+\sqrt2=3+4\sqrt2$
$a_6=a_5+d = 3+4\sqrt2+\sqrt2=3+5\sqrt2$
$a_7=a_6+d = 3+5\sqrt2+\sqrt2=3+6\sqrt2$
Therefore, next three terms of given series are $3+4\sqrt2, 3+5\sqrt2,3+6\sqrt2$

$\small 0.2,0.22,0.222,0.2222,...$

Given series is
$\small 0.2,0.22,0.222,0.2222,...$
Now,
the first term to this series is = 0.2
Now,
$a_1 = 0.2 \ \ and \ \ a_2 = 0.22 \ \ and \ \ a_3 = 0.222 \ \ and \ \ a_4 = 0.2222$
$a_2-a_1 = 0.22-0.2=0.02$
$a_3-a_2 = 0.222-0.22=0.002$

We can clearly see that the difference between terms are not equal
Hence, given series is not an AP

$\small 0,-4,-8,-12,...$

Given series is
$\small 0,-4,-8,-12,...$
Now,
first term to this series is = 0
Now,
$a_1 = 0 \ \ and \ \ a_2 = -4 \ \ and \ \ a_3 = -8 \ \ and \ \ a_4 = -12$
$a_2-a_1 = -4-0 =-4$
$a_3-a_2 = -8-(-4) =-8+4 = -4$
$a_4-a_3=-12-(-8)=-12+8=-4$
We can clearly see that the difference between terms are equal and equal to -4
Hence, given series is in AP
Now, the next three terms are
$a_5=a_4+d = -12-4 =-16$
$a_6=a_5+d = -16-4=-20$
$a_7=a_6+d = -20-4=-24$
Therefore, the next three terms of given series are -16,-20,-24

Q4 (viii) Which of the following are APs? If they form an AP, find the common difference d and write three more terms. $\small -\frac{1}{2},-\frac{1}{2},-\frac{1}{2},-\frac{1}{2},...$

Given series is
$\small -\frac{1}{2},-\frac{1}{2},-\frac{1}{2},-\frac{1}{2},...$
Now,
the first term to this series is = $-\frac{1}{2}$
Now,
$a_1 = -\frac{1}{2} \ \ and \ \ a_2 = -\frac{1}{2} \ \ and \ \ a_3 = -\frac{1}{2} \ \ and \ \ a_4 = -\frac{1}{2}$
$a_2-a_1 = -\frac{1}{2}-\left ( -\frac{1}{2} \right ) = -\frac{1}{2}+\frac{1}{2}=0$
$a_3-a_2 = -\frac{1}{2}-\left ( -\frac{1}{2} \right ) = -\frac{1}{2}+\frac{1}{2}=0$
$a_4-a_3 = -\frac{1}{2}-\left ( -\frac{1}{2} \right ) = -\frac{1}{2}+\frac{1}{2}=0$
We can clearly see that the difference between terms are equal and equal to 0
Hence, given series is in AP
Now, the next three terms are
$a_5=a_4+d = -\frac{1}{2}+0=-\frac{1}{2}$
$a_6=a_5+d = -\frac{1}{2}+0=-\frac{1}{2}$
$a_7=a_6+d = -\frac{1}{2}+0=-\frac{1}{2}$
Therefore, the next three terms of given series are $-\frac{1}{2},-\frac{1}{2},-\frac{1}{2}$

$\small 1,3,9,27,...$

Given series is
$\small 1,3,9,27,...$
Now,
the first term to this series is = 1
Now,
$a_1 = 1 \ \ and \ \ a_2 = 3 \ \ and \ \ a_3 = 9 \ \ and \ \ a_4 = 27$
$a_2-a_1 = 3-1=2$
$a_3-a_2 =9-3=6$

We can clearly see that the difference between terms are not equal
Hence, given series is not an AP

$\small a,2a,3a,4a,...$

Given series is
$\small a,2a,3a,4a,...$
Now,
the first term to this series is = a
Now,
$a_1 = a \ \ and \ \ a_2 = 2a \ \ and \ \ a_3 = 3a \ \ and \ \ a_4 = 4a$
$a_2-a_1 = 2a-a =a$
$a_3-a_2 = 3a-2a =a$
$a_4-a_3=4a-3a=a$
We can clearly see that the difference between terms are equal and equal to a
Hence, given series is in AP
Now, the next three terms are
$a_5=a_4+d =4a+a=5a$
$a_6=a_5+d =5a+a=6a$
$a_7=a_6+d =6a+a=7a$
Therefore, next three terms of given series are 5a,6a,7a

$\small a,a^2,a^3,a^4,...$

Given series is
$\small a,a^2,a^3,a^4,...$
Now,
the first term to this series is = a
Now,
$a_1 = a \ \ and \ \ a_2 = a^2 \ \ and \ \ a_3 = a^3 \ \ and \ \ a_4 = a^4$
$a_2-a_1 = a^2-a =a(a-1)$
$a_3-a_2 = a^3-a^2 =a^2(a-1)$

We can clearly see that the difference between terms are not equal
Hence, given series is not in AP

$\small \sqrt{2},\sqrt{8},\sqrt{18},\sqrt{32},...$

Given series is
$\small \sqrt{2},\sqrt{8},\sqrt{18},\sqrt{32},...$
We can rewrite it as
$\sqrt2,2\sqrt2,3\sqrt2,4\sqrt2,....$
Now,
first term to this series is = a
Now,
$a_1 = \sqrt2 \ \ and \ \ a_2 = 2\sqrt2 \ \ and \ \ a_3 = 3\sqrt2 \ \ and \ \ a_4 = 4\sqrt2$
$a_2-a_1 = 2\sqrt2-\sqrt2 =\sqrt2$
$a_3-a_2 = 3\sqrt2-2\sqrt2 =\sqrt2$
$a_4-a_3=4\sqrt2-3\sqrt2=\sqrt2$
We can clearly see that difference between terms are equal and equal to $\sqrt2$
Hence, given series is in AP
Now, the next three terms are
$a_5=a_4+d =4\sqrt2+\sqrt2=5\sqrt2$
$a_6=a_5+d =5\sqrt2+\sqrt2=6\sqrt2$
$a_7=a_6+d =6\sqrt2+\sqrt2=7\sqrt2$
Therefore, next three terms of given series are $5\sqrt2,6\sqrt2,7\sqrt2$

That is the next three terms are $\sqrt{50},\ \sqrt{72},\ \sqrt{98}$

$\small \sqrt{3},\sqrt{6},\sqrt{9},\sqrt{12},...$

Given series is
$\small \sqrt{3},\sqrt{6},\sqrt{9},\sqrt{12},...$
Now,
the first term to this series is = $\sqrt3$
Now,
$a_1 = \sqrt3 \ \ and \ \ a_2 = \sqrt6 \ \ and \ \ a_3 = \sqrt9 \ \ and \ \ a_4 = \sqrt{12}$
$a_2-a_1 = \sqrt6-\sqrt3 =\sqrt3(\sqrt2-1)$
$a_3-a_2 = 3-\sqrt3 =\sqrt3(\sqrt3-1)$
We can clearly see that the difference between terms are not equal
Hence, given series is not in AP

$\small 1^2,3^2,5^2,7^2,...$

Given series is
$\small 1^2,3^2,5^2,7^2,...$
we can rewrite it as
$1,9,25,49,....$
Now,
the first term to this series is = 1
Now,
$a_1 =1 \ \ and \ \ a_2 = 9 \ \ and \ \ a_3 =25 \ \ and \ \ a_4 = 49$
$a_2-a_1 = 9-1 = 8$
$a_3-a_2 = 25-9=16$
We can clearly see that the difference between terms are not equal
Hence, given series is not in AP

Given series is
$\small 1^2,5^2,7^2,73,...$
we can rewrite it as
$1,25,49,73....$
Now,
the first term to this series is = 1
Now,
$a_1 =1 \ \ and \ \ a_2 = 25 \ \ and \ \ a_3 =49 \ \ and \ \ a_4 = 73$
$a_2-a_1 = 25-1 = 24$
$a_3-a_2 = 49-25=24$
$a_4-a_3 = 73-49=24$
We can clearly see that the difference between terms are equal and equal to 24
Hence, given series is in AP
Now, the next three terms are
$a_5=a_4+d = 73+24=97$
$a_6=a_5+d = 97+24=121$
$a_7=a_6+d = 121+24=145$
Therefore, the next three terms of given series are 97,121,145

## More About NCERT Solutions for Class 10 Maths Exercise 5.2

The problems in exercise 5.1 for Class 10 Maths are divided into four categories. Firstly we have a very basic question we have been given scenarios and we have to tell whether it is A.P or not then we have direct questions in which we have been given first term and common difference and we have to find the A.P series NCERT solutions for Class 10 Maths exercise 5.1 also consist of question in which we have to find first term and common difference and also we have question-based on finding whether series is AP or not and if it has to write more terms. All sorts of fundamental problems based on the AP general equation, first term, and common difference are covered in exercise 5.1 Class 10 Maths. Students can quickly go through the Arithmetic Progressions Class 10 Notes to revise all concepts all together.

## Benefits of NCERT Solutions for Class 10 Maths Exercise 5.1

• Class 10 Mathematics chapter 5 exercise 5.1 covers a broad range of topics that will be covered in greater depth in Class 10 Maths chapter 5 exercise 5.2.
• NCERT Class 10 Maths chapter 5 exercise 5.1, will be helpful in Class 11 Maths chapter 9 sequences and series
• Exercise 5.1 Class 10 Maths, is based on the general idea about arithmetic progression, which is important for building other concepts of the chapter.

Also see-

## Subject Wise NCERT Exemplar Solutions

1. What do you mean by arithmetic progression according to NCERT solutions for Class 10 Maths 5 exercise 5.1?

Arithmetic progression is a series of numbers in which each term is obtained by adding a constant number to the next term except the first term.

2. What is general equation A.P series according to NCERT solutions for Class 10 Maths 5 exercise 5.1?

The general equation of the A.P series is a, a+d, a+2d and so on.

3. Give an example of A.P series

A very basic example of A.P series is 1,2,3,4 in which the first term is 1 and the common difference is also 1.

4. According to NCERT solutions for Class 10 Maths 5 exercise 5.1 what can be the value of common difference ?

Common differences can be zero, positive and even negative.

5. Is 2,2,2,2 is considered an A.P series.

Yes, it is considered A.P series as in exercise 5.1 Class 10 Maths it is clearly stated common difference can be zero so it is an AP series with a common difference zero.

6. What is the difference between finite and infinite A.P series?

The finite A.P series has limited-term while the infinite A.P series has unlimited terms and it never ends.

7. What is the number of solved examples before the exercise 5.1 Class 10 Maths?

Before the Class 10 Mathematics chapter 5 exercise 5.1, there are two primary questions that must be answered.

8. In the NCERT answers for Class 10 Maths chapter 5 exercise 5.1, how many questions are covered?

There are 4 questions in Class 10th Maths chapter exercise 5.1. Question 1 consist of four subparts.

Question 2 consists of 5 subparts. Question 3 consists of four subparts. Question 1 consists of 15 subparts

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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.

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) 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