NCERT Solutions for Exercise 10.3 Class 12 Maths Chapter 10

NCERT Solutions for Exercise 10.3 Class 12 Maths Chapter 10 - Vector Algebra

Edited By Ramraj Saini | Updated on Dec 04, 2023 08:54 AM IST | #CBSE Class 12th

NCERT Solutions For Class 12 Maths Chapter 10 Exercise 10.3

NCERT Solutions for Exercise 10.3 Class 12 Maths Chapter 10 Vector Algebra are discussed here. These NCERT solutions are created by subject matter expert at Careers360 considering the latest syllabus and pattern of CBSE 2023-24. NCERT solutions for exercise 10.3 Class 12 Maths chapter 10 explains the questions related to dot products of vectors. As far as the unit vector algebra is concerned dot product is an important topic. NCERT syllabus Exercise 10.3 Class 12 Math familiarise the students with the concept of dot products. NCERT solutions for Class 12 Maths chapter 10 exercise 10.3 are framed by maths expert and is in accordance with the CBSE syllabus. Students can use Class 12 Maths chapter 10 exercise 10.3 for the preparations of CBSE Class 12 Board Exams. Along with NCERT practice CBSE Previous Year Class 12 Maths Questions to get a good score in board exam.

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12th class Maths exercise 10.3 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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Vector Algebra Class 12 Chapter 10 Exercise: 10.3

Question:1 Find the angle between two vectors $\vec a \: \:and \: \: \vec b$ with magnitudes $\sqrt 3 \: \:and \: \: 2$ , respectively having . $\vec a . \vec b = \sqrt 6$

Answer:

Given

$\left | \vec a \right |=\sqrt{3}$

$\left | \vec b \right |=2$

$\vec a . \vec b = \sqrt 6$

As we know

$\vec a . \vec b = \left | \vec a \right |\left | \vec b \right |cos\theta$

where $\theta$ is the angle between two vectors

So,

$cos\theta =\frac{\vec a.\vec b}{\left | \vec a \right |\left | \vec b \right |}=\frac{\sqrt{6}}{\sqrt{3}*2}=\frac{1}{\sqrt{2}}$

$\theta=\frac{\pi}{4}$

Hence the angle between the vectors is $\frac{\pi}{4}$ .

Question:2 Find the angle between the vectors $\hat i - 2 \hat j + 3 \hat k \: \:and \: \: 3 \hat i - 2 \hat j + \hat k$

Answer:

Given two vectors

$\vec a=\hat i - 2 \hat j + 3 \hat k \: \:and \: \: \vec b=3 \hat i - 2 \hat j + \hat k$

Now As we know,

The angle between two vectors $\vec a$ and $\vec b$ is given by

$\theta=cos^{-1}\left ( \frac{\vec a.\vec b}{\left | \vec a \right |\left | \vec b \right |}\right )$

Hence the angle between $\vec a=\hat i - 2 \hat j + 3 \hat k \: \:and \: \: \vec b=3 \hat i - 2 \hat j + \hat k$

$\theta=cos^{-1}\left ( \frac{(\hat i-2\hat j+3\hat k).(3\hat i-2\hat j+\hat k)}{\left | \hat i-2\hat j+3\hat k \right |\left |3\hat i-2\hat j+\hat k \right |}\right )$

$\theta=cos^{-1}\left ( \frac{3+4+3}{\sqrt{1^2+(-2)^2+3^3}\sqrt{3^2+(-2)^2+1^2}} \right )$

$\theta=cos^{-1}\frac{10}{14}$

$\theta=cos^{-1}\frac{5}{7}$

Question:3 Find the projection of the vector $\hat i - \hat j$ on the vector $\hat i + \hat j$

Answer:

Let

$\vec a=\hat i - \hat j$

$\vec b=\hat i + \hat j$

Projection of vector $\vec a$ on $\vec b$

$\frac{\vec a.\vec b}{\left | \vec b \right |}=\frac{(\hat i-\hat j)(\hat i+\hat j)}{\left |\hat i+\hat j \right |}=\frac{1-1}{\sqrt{2}}=0$

Hence, Projection of vector $\vec a$ on $\vec b$ is 0.

Question:4 Find the projection of the vector $\hat i + 3 \hat j + 7 \hat k$ on the vector $7\hat i - \hat j + 8 \hat k$

Answer:

Let

$\vec a =\hat i + 3 \hat j + 7 \hat k$

$\vec b=7\hat i - \hat j + 8 \hat k$

The projection of $\vec a$ on $\vec b$ is

$\frac{\vec a.\vec b}{\left | \vec b \right |}=\frac{(\hat i+3\hat j+7\hat k)(7\hat i-\hat j+8\hat k)}{\left | 7\hat i-\hat j+8\hat k \right |}=\frac{7-3+56}{\sqrt{7^2+(-1)^2+8^2}}=\frac{60}{\sqrt{114}}$

Hence, projection of vector $\vec a$ on $\vec b$ is

$\frac{60}{\sqrt{114}}$

Question:5 Show that each of the given three vectors is a unit vector: $\frac{1}{7}( 2 \hat i + 3 \hat j + 6 \hat k ), \frac{1}{7}( 3 \hat i- 6 \hat j + 2 \hat k ), \frac{1}{7}( 6\hat i + 2 \hat j -3\hat k )$ Also, show that they are mutually perpendicular to each other.

Answer:

Given

$\\\vec a=\frac{1}{7}( 2 \hat i + 3 \hat j + 6 \hat k ), \\\ \vec b =\frac{1}{7}( 3 \hat i- 6 \hat j + 2 \hat k ),\\\vec c = \frac{1}{7}( 6\hat i + 2 \hat j -3\hat k )$

Now magnitude of $\vec a,\vec b \:and\: \vec c$

$\left | \vec a \right |=\frac{1}{7} \sqrt{2^2+3^2+6^2}=\frac{\sqrt{49}}{7}=1$

$\left | \vec b \right |=\frac{1}{7} \sqrt{3^2+(-6)^2+2^2}=\frac{\sqrt{49}}{7}=1$

$\left | \vec c \right |=\frac{1}{7} \sqrt{6^2+2^2+(-3)^2}=\frac{\sqrt{49}}{7}=1$

Hence, they all are unit vectors.

Now,

$\vec a.\vec b=\frac{1}{7}(2\hat i+3\hat j+6\hat k)\frac{1}{7}(3\hat i-6\hat j+2\hat k)=\frac{1}{49}(6-18+12)=0$

$\vec b.\vec c=\frac{1}{7}(3\hat i-6\hat j+2\hat k)\frac{1}{7}(6\hat i+2\hat j-3\hat k)=\frac{1}{49}(18-12-6)=0$

$\vec c.\vec a=\frac{1}{7}(6\hat i+2\hat j-3\hat k)\frac{1}{7}(2\hat i+3\hat j-6\hat k)=\frac{1}{49}(12+6-18)=0$

Hence all three are mutually perpendicular to each other.

Answer:

Given in the question

$( \vec a + \vec b ). ( \vec a - \vec b )=8$

$\left | \vec a \right |^2-\left | \vec b \right |^2=8$

Since $|\vec a |\: \:= 8 \: \:|\vec b |$

$\left | \vec {8b} \right |^2-\left | \vec b \right |^2=8$

$\left | \vec {63b} \right |^2=8$

$\left | \vec {b} \right |^2=\frac{8}{63}$

$\left | \vec {b} \right |=\sqrt{\frac{8}{63}}$

So, answer of the question is

$\left | \vec {a} \right |=8\left | \vec {b} \right |=8\sqrt{\frac{8}{63}}$

Question:7 Evaluate the product $( 3\vec a - 5 \vec b ). ( 2 \vec a + 7 \vec b )$ .

Answer:

To evaluate the product $( 3\vec a - 5 \vec b ). ( 2 \vec a + 7 \vec b )$

$( 3\vec a - 5 \vec b ). ( 2 \vec a + 7 \vec b )=6\vec a.\vec a+21\vec a.\vec b-10\vec b.\vec a-35\vec b.\vec b$

$=6\vec a.^2+11\vec a.\vec b-35\vec b^2$

$=6\left | \vec a \right |^2+11\vec a.\vec b-35\left | \vec b \right |^2$

Question:8 Find the magnitude of two vectors $\vec a \: \: and \: \: \vec b$ , having the same magnitude and such that the angle between them is $60 \degree$ and their scalar product is 1/2

Answer:

Given two vectors $\vec a \: \: and \: \: \vec b$

$\left | \vec a \right |=\left | \vec b\right |$

$\vec a.\vec b=\frac{1}{2}$

Now Angle between $\vec a \: \: and \: \: \vec b$

$\theta=60^0$

Now As we know that

$\vec a.\vec b=\left | \vec a \right |\left | \vec b \right |cos\theta$

$\frac{1}{2}=\left | \vec a \right |\left | \vec a \right |cos60^0$

$\left | a \right |^2=1$

Hence, the magnitude of two vectors $\vec a \: \: and \: \: \vec b$

$\left | a \right |=\left | b \right |=1$

Question:9 Find $|\vec x |$ , if for a unit vector $\vec a , ( \vec x -\vec a ) . ( \vec x + \vec a ) = 12$

Answer:

Given in the question that

$( \vec x -\vec a ) . ( \vec x + \vec a ) = 12$

And we need to find $\left | \vec x \right |$

$\left | \vec x \right |^2-\left | \vec a \right |^2 = 12$

$\left | \vec x \right |^2-1 = 12$

$\left | \vec x \right |^2 = 13$

$\left | \vec x \right | = \sqrt{13}$

So the value of $\left | \vec x \right |$ is $\sqrt{13}$

Question:10 If $\vec a = 2 \hat i + 2 \hat j + 3 \hat k , \vec b = - \hat i + 2 \hat j + \hat k \: \: and \: \: \vec c = 3 \hat i + \hat j$ are such that $\vec a + \lambda \vec b$ is perpendicular to $\vec c$ , then find the value of $\lambda$

Answer:

Given in the question is

$\vec a = 2 \hat i + 2 \hat j + 3 \hat k , \vec b = - \hat i + 2 \hat j + \hat k \: \: and \: \: \vec c = 3 \hat i + \hat j$

and $\vec a + \lambda \vec b$ is perpendicular to $\vec c$

and we need to find the value of $\lambda$ ,

so the value of $\vec a + \lambda \vec b$ -

$\vec a + \lambda \vec b=2\hat i +2\hat j +3\hat k+\lambda (-\hat i+2\hat j+\hat k)$

$\vec a + \lambda \vec b=(2-\lambda)\hat i +(2+2\lambda)\hat j +(3+\lambda)\hat k$

As $\vec a + \lambda \vec b$ is perpendicular to $\vec c$

$(\vec a + \lambda \vec b).\vec c=0$

$((2-\lambda)\hat i +(2+2\lambda)\hat j +(3+\lambda)\hat k)(3\hat i+\hat j)=0$

$3(2-\lambda)+2+2\lambda=0$

$6-3\lambda+2+2\lambda=0$

$\lambda=8$

the value of $\lambda=8$ ,

Answer:

Given in the question that -

$\vec a \: \: \: and \: \: \vec b$ are two non-zero vectors

According to the question

$=|\vec a |^2 |\vec b|^2 - |\vec b |^2 |\vec a|^2+|\vec b||\vec a|\vec a.\vec b-|\vec a||\vec b|\vec b.\vec a=0$

Question:12 If $\vec a . \vec a = 0 \: \: and \: \: \vec a . \vec b = 0$ , then what can be concluded about the vector $\vec b$ ?

Answer:

Given in the question

$\\\vec a . \vec a = 0 \\|\vec a|^2=0$

$\\|\vec a|=0$

Therefore $\vec a$ is a zero vector. Hence any vector $\vec b$ will satisfy $\vec a . \vec b = 0$

Question:13 If $\vec a , \vec b , \vec c$ are unit vectors such that $\vec a + \vec b + \vec c = \vec 0$ , find the value of $\vec a . \vec b + \vec b. \vec c + \vec c . \vec a$

Answer:

Given in the question

$\vec a , \vec b , \vec c$ are unit vectors $\Rightarrow |\vec a|=|\vec b|=|\vec c|=1$

and $\vec a + \vec b + \vec c = \vec 0$

and we need to find the value of $\vec a . \vec b + \vec b. \vec c + \vec c . \vec a$

$(\vec a + \vec b + \vec c)^2 = \vec 0$

$\vec a^2 + \vec b^2 + \vec c ^2+2(\vec a . \vec b + \vec b. \vec c + \vec c . \vec a)=0$

$|\vec a|^2 + |\vec b|^2 + |\vec c |^2+2(\vec a . \vec b + \vec b. \vec c + \vec c . \vec a)=0$

$1+1+1+2(\vec a . \vec b + \vec b. \vec c + \vec c . \vec a)=0$

$\vec a . \vec b + \vec b. \vec c + \vec c . \vec a=\frac{-3}{2}$

Answer- the value of $\vec a . \vec b + \vec b. \vec c + \vec c . \vec a$ is $\frac{-3}{2}$

Question:14 If either vector $\vec a = 0 \: \: or \: \: \vec b = 0 \: \: then \: \: \vec a . \vec b = 0$ . But the converse need not be true. Justify your answer with an example

Answer:

Let

$\vec a=\hat i-2\hat j +3\hat k$

$\vec b=5\hat i+4\hat j +1\hat k$

we see that

$\vec a.\vec b=(\hat i-2\hat j +3\hat k)(5\hat i+4\hat j +1\hat k)=5-8+3=0$

we now observe that

$|\vec a|=\sqrt{1^2+(-2)^2+3^2}=\sqrt{14}$

$|\vec b|=\sqrt{5^2+4^2+1^2}=\sqrt{42}$

Hence here converse of the given statement is not true.

Question:15 If the vertices A, B, C of a triangle ABC are (1, 2, 3), (–1, 0, 0), (0, 1, 2), respectively, then find $\angle ABC , [\angle ABC$ is the angle between the vectors $\overline{BA}\: \: and\: \: \overline{BC} ]$ .

Answer:

Given points,

A=(1, 2, 3),

B=(–1, 0, 0),

C=(0, 1, 2),

As need to find Angle between $\overline{BA}\: \: and\: \: \overline{BC} ]$

$\vec {BA}=(1-(-1))\hat i+(2-0)\hat j+(3-0)\hat k=2\hat i+2\hat j+3\hat k$

$\vec {BC}=(0-(-1))\hat i+(1-0)\hat j+(2-0)\hat k=\hat i+\hat j+2\hat k$

Hence angle between them ;

$\theta=cos^{-1}(\frac{\vec {BA}.\vec {BC}}{\left | \vec {BA} \right |\left | \vec {BC} \right |})$

$\theta=cos^{-1}\frac{2+2+6}{\sqrt{17}\sqrt{6}}$

$\theta=cos^{-1}\frac{10}{\sqrt{102}}$

Answer - Angle between the vectors $\overline{BA}\: \: and\: \: \overline{BC}$ is $\theta=cos^{-1}\frac{10}{\sqrt{102}}$

Question:16 Show that the points A(1, 2, 7), B(2, 6, 3) and C(3, 10, –1) are collinear.

Answer:

Given in the question

A=(1, 2, 7), B=(2, 6, 3) and C(3, 10, –1)

To show that the points A(1, 2, 7), B(2, 6, 3) and C(3, 10, –1) are collinear

$\vec {AB}=(2-1)\hat i+(6-2)\hat j+(3-7)\hat k$

$\vec {AB}=\hat i+4\hat j-4\hat k$

$\vec {BC}=(3-2)\hat i+(10-6)\hat j+(-1-3)\hat k$

$\vec {BC}=\hat i+4\hat j-4\hat k$

$\vec {AC}=(3-1)\hat i+(10-2)\hat j+(-1-7)\hat k$

$\vec {AC}=2\hat i+8\hat j-8\hat k$

$|\vec {AB}|=\sqrt{1^2+4^2+(-4)^2}=\sqrt{33}$

$|\vec {BC}|=\sqrt{1^2+4^2+(-4)^2}=\sqrt{33}$

$|\vec {AC}|=\sqrt{2^2+8^2+(-8)^2}=2\sqrt{33}$

As we see that

$|\vec {AC}|=|\vec {AB}|+|\vec {BC}|$

Hence point A, B , and C are colinear.

Question:17 Show that the vectors $2 \hat i - \hat j + \hat k , \hat i - 3 \hat j - 5 \hat k \: \: and \: \: 3 \hat i - 4 \hat j - 4 \hat k$ form the vertices of a right angled triangle.

Answer:

Given the position vector of A, B , and C are

$2 \hat i - \hat j + \hat k , \hat i - 3 \hat j - 5 \hat k \: \: and \: \: 3 \hat i - 4 \hat j - 4 \hat k$

To show that the vectors $2 \hat i - \hat j + \hat k , \hat i - 3 \hat j - 5 \hat k \: \: and \: \: 3 \hat i - 4 \hat j - 4 \hat k$ form the vertices of a right angled triangle

$\vec {AB}=(1-2)\hat i + (-3-(-1))\hat j+(-5-1)\hat k=-1\hat i -2\hat j-6\hat k$

$\vec {BC}=(3-1)\hat i + (-4-(-3))\hat j+(-4-(-5))\hat k=-2\hat i -\hat j+\hat k$

$\vec {AC}=(3-2)\hat i + (-4-(-1))\hat j+(-4-(1))\hat k=\hat i -3\hat j-5\hat k$

$|\vec {AB}|=\sqrt{(-1)^2+(-2)^2+(-6)^2}=\sqrt{41}$

$|\vec {BC}|=\sqrt{(-2)^2+(-1)^2+(1)^2}=\sqrt{6}$

$|\vec {AC}|=\sqrt{(1)^2+(-3)^2+(-5)^2}=\sqrt{35}$

Here we see that

$|\vec {AC}|^2+|\vec {BC}|^2=|\vec {AB}|^2$

Hence A,B, and C are the vertices of a right angle triangle.

Question:18 If $\vec a$ is a nonzero vector of magnitude ‘a’ and $\lambda$ a nonzero scalar, then $\lambda \vec a$ is unit vector if

$\\A ) \lambda = 1 \\\\ B ) \lambda = -1 \\\\ C ) a = |\lambda | \\\\ D ) a = 1 / |\lambda |$

Answer:

Given $\vec a$ is a nonzero vector of magnitude ‘a’ and $\lambda$ a nonzero scalar

$\lambda \vec a$ is a unit vector when

$|\lambda \vec a|=1$

$|\lambda|| \vec a|=1$

$| \vec a|=\frac{1}{|\lambda|}$

Hence the correct option is D.

Benefits of NCERT Solutions for Class 12 Maths Chapter 10 Exercise 10.3

Similar questions as in Class 12th Maths chapter 10 exercise 10.3 can be expected for the CBSE Class 12 Maths exam. The questions may not be exactly the same.
Problems based on dot products are also asked in various competitive exams like JEE Main, KEAM etc.
Problems discussed in exercise 10.3 Class 12 Maths are useful not only for Class 12 Maths but also for Class 11 and 12 Physics also.

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Key Features Of NCERT Solutions for Exercise 10.3 Class 12 Maths Chapter 10

Comprehensive Coverage: The solutions encompass all the topics covered in ex 10.3 class 12, ensuring a thorough understanding of the concepts.
Step-by-Step Solutions: In this class 12 maths ex 10.3, each problem is solved systematically, providing a stepwise approach to aid in better comprehension for students.
Accuracy and Clarity: Solutions for class 12 ex 10.3 are presented accurately and concisely, using simple language to help students grasp the concepts easily.
Conceptual Clarity: In this 12th class maths exercise 10.3 answers, emphasis is placed on conceptual clarity, providing explanations that assist students in understanding the underlying principles behind each problem.
Inclusive Approach: Solutions for ex 10.3 class 12 cater to different learning styles and abilities, ensuring that students of various levels can grasp the concepts effectively.
Relevance to Curriculum: The solutions for class 12 maths ex 10.3 align closely with the NCERT curriculum, ensuring that students are prepared in line with the prescribed syllabus.

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Frequently Asked Question (FAQs)

1. Is it possible to have two vectors whose magnitude is non zero but their dot product is zero?

Yes. For two perpendicular vectors, the dot product is zero.

2. What are the main topics covered in the chapter vector algebra?

The main three topics are the addition of vectors, the dot product of vectors and the cross product of vectors.

3. Why work is a scalar quantity even though force and displacement are vectors?

The work done is the dot product of force and displacement. The dot product of two vectors is a scalar (real number).

4. What is the quantity obtained from the dot product of force and velocity?

The dot product of force and velocity gives power.

5. Why i.k=0?

i.k=0 as the angle between them is 90 degrees

6. Can dot product be a negative number?

Yes, the dot product of two vectors can be either positive, negative or zero based on the angle between them.

7. How many exercises are solved in NCERT Class 12 chapter vector algebra?

There are a total of 5 exercises including miscellaneous.

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Questions related to CBSE Class 12th

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i upload 12th board result screenshot in ncweb form from the cbse site so it is valid

hello mahima,

If you have uploaded screenshot of your 12th board result taken from CBSE official website,there won,t be a problem with that.If the screenshot that you have uploaded is clear and legible. It should display your name, roll number, marks obtained, and any other relevant details in a readable forma.ALSO, the screenshot clearly show it is from the official CBSE results portal.

hope this helps.

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Ashvadha 12 July,2024

Kya mujhe class 12 ka important question milega jo exam mai per year puche jate hai

Hello Akash,

If you are looking for important questions of class 12th then I would like to suggest you to go with previous year questions of that particular board. You can go with last 5-10 years of PYQs so and after going through all the questions you will have a clear idea about the type and level of questions that are being asked and it will help you to boost your class 12th board preparation.

You can get the Previous Year Questions (PYQs) on the official website of the respective board.

I hope this answer helps you. If you have more queries then feel free to share your questions with us we will be happy to assist you.

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kumardurgesh1802 10 July,2024

I have completed my 12th with arts cbse. I want to repeat my 12th with pcmb. How to become a private candidate and what are deadline to be able to appear in 2025 exam?

Hello student,

If you are planning to appear again for class 12th board exam with PCMB as a private candidate here is the right information you need:

No school admission needed! Register directly with CBSE. (But if you want to attend the school then you can take admission in any private school of your choice but it will be waste of money)
You have to appear for the 2025 12th board exams.
Registration for class 12th board exam starts around September 2024 (check CBSE website for exact dates).
Aim to register before late October to avoid extra fees.
Schools might not offer classes for private students, so focus on self-study or coaching.

Remember , these are tentative dates based on last year. Keep an eye on the CBSE website ( https://www.cbse.gov.in/ ) for the accurate and official announcement.

I hope this answer helps you. If you have more queries then feel free to share your questions with us, we will be happy to help you.

Good luck with your studies!

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kumardurgesh1802 10 July,2024

I passed my 12th cbse boards in 2023 , and first time appear for jee mains and advance in 2024 so , am i eligible for 2025 jee mains and advance ?

Hello Aspirant , Hope your doing great . As per your query , your eligible for JEE mains in the year of 2025 , Every candidate can appear for the JEE Main exam 6 times over three consecutive years . The JEE Main exam is held two times every year, in January and April.

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satyamercedes0626 23 July,2024

I've passed my class 12th board exam from cbse in PCB in 2024. Now I'll give mathematics exam from cbse as a additional subject in 2025. However, they'll provide separate marksheet for maths. Will i be eligible for jee?? And how I'll fill the form??

Hi there,

Hope you are doing fine

Yes you are certainly eligible for giving the jee exam in the year 2025. You must pass the maths exam with at least 75% criteria as required by jee and provide the marksheet and the passing certificate while registering for the exam.

Pursuing maths as an additional subject while taking biology as your main subject does not offer any hindrance in you appearing for the jee examination. It is indeed an privilege to pursue both maths and biology as the subjects and prepare for the same.

There will be no issue in filling the form while registering for the exam as it will only require your basic details and marksheet which you can provide by attaching the marksheet of maths also. Also, a detailed roadmap is also available on the official websites on how to fill the registration form. So you can fill the form easily.

Hope this resolves your query.

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Abhishek 12 July,2024

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Option 1) $0.34\; J$	Option 2) $0.16\; J$
Option 3) $1.00\; J$	Option 4) $0.67\; J$

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) $20,000 \, \, J - 50,000 \, \, J$

Option 1) $K/2\,$	Option 2) $\; K\;$
Option 3) $zero\;$	Option 4) $K/4$

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 .

Option 1) 0.02	Option 2) 3.125 × 10^-2
Option 3) 1.25 × 10^-2	Option 4) 2.5 × 10^-2

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.

Option 1) Molality	Option 2) Weight fraction of solute
Option 3) Fraction of solute present in water	Option 4) Mole fraction.

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

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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