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

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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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NCERT Solutions Subject Wise

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Frequently Asked Questions (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

Have a question related to CBSE Class 12th ?

Can I change my board from CBSE to CHSE Odisha in Class 12th?

Hello,

Yes, you can switch from CBSE to CHSE Odisha in Class 12th, but there are a few conditions:

Eligibility : Check if CHSE Odisha allows lateral entry into Class 12th. Generally, students can transfer from CBSE to state boards with prior approval.
Document Requirements : You may need to provide transcripts, mark sheets, and a transfer certificate.
Subject Alignment : Ensure that the subjects you studied in CBSE align with the CHSE curriculum.
Application Process : Contact the respective CHSE Odisha authority or school for specific transfer procedures and deadlines.

Hope it helps !

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Prachi Kumari 30 December,2024

quartely question paper for mathematics cbse

Hello there! Thanks for reaching out to us at Careers360.

Ah, you're looking for CBSE quarterly question papers for mathematics, right? Those can be super helpful for exam prep.

Unfortunately, CBSE doesn't officially release quarterly papers - they mainly put out sample papers and previous years' board exam papers. But don't worry, there are still some good options to help you practice!

Have you checked out the CBSE sample papers on their official website? Those are usually pretty close to the actual exam format. You could also look into previous years' board exam papers - they're great for getting a feel for the types of questions that might come up.

If you're after more practice material, some textbook publishers release their own mock papers which can be useful too.

Let me know if you need any other tips for your math prep. Good luck with your studies!

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Jefferson 25 September,2024

i have taken admission in clg after my campartment exam in chemistry along with the neet 2025 preperation but unfortunately i again got campartment after doing all this preparation i got 15 marks only now I can't able to think what should I do next.

It's understandable to feel disheartened after facing a compartment exam, especially when you've invested significant effort. However, it's important to remember that setbacks are a part of life, and they can be opportunities for growth.

Possible steps:

Re-evaluate Your Study Strategies:
- Identify Weak Areas: Pinpoint the specific topics or concepts that caused difficulties.
- Seek Clarification: Reach out to teachers, tutors, or online resources for additional explanations.
- Practice Regularly: Consistent practice is key to mastering chemistry.
Consider Professional Help:
- Tutoring: A tutor can provide personalized guidance and support.
- Counseling: If you're feeling overwhelmed or unsure about your path, counseling can help.
Explore Alternative Options:
- Retake the Exam: If you're confident in your ability to improve, consider retaking the chemistry compartment exam.
- Change Course: If you're not interested in pursuing chemistry further, explore other academic options that align with your interests.
Focus on NEET 2025 Preparation:
- Stay Dedicated: Continue your NEET preparation with renewed determination.
- Utilize Resources: Make use of study materials, online courses, and mock tests.
Seek Support:
- Talk to Friends and Family: Sharing your feelings can provide comfort and encouragement.
- Join Study Groups: Collaborating with peers can create a supportive learning environment.

Remember: This is a temporary setback. With the right approach and perseverance, you can overcome this challenge and achieve your goals.

I hope this information helps you.

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Kanishka kaushikii 17 September,2024

i scored 84 percent in cbse 2 years ago.Am i eligible for medhavi scholarship if give 12th again after 2 drop years and will got 75+ in mpboard

Hi,

Qualifications:
Age: As of the last registration date, you must be between the ages of 16 and 40.
Qualification: You must have graduated from an accredited board or at least passed the tenth grade. Higher qualifications are also accepted, such as a diploma, postgraduate degree, graduation, or 11th or 12th grade.
How to Apply:
Get the Medhavi app by visiting the Google Play Store.
Register: In the app, create an account.
Examine Notification: Examine the comprehensive notification on the scholarship examination.
Sign up to Take the Test: Finish the app's registration process.
Examine: The Medhavi app allows you to take the exam from the comfort of your home.
Get Results: In just two days, the results are made public.
Verification of Documents: Provide the required paperwork and bank account information for validation.
Get Scholarship: Following a successful verification process, the scholarship will be given. You need to have at least passed the 10th grade/matriculation scholarship amount will be transferred directly to your bank account.

Scholarship Details:

Type A: For candidates scoring 60% or above in the exam.

Type B: For candidates scoring between 50% and 60%.

Type C: For candidates scoring between 40% and 50%.

Cash Scholarship:

Scholarships can range from Rs. 2,000 to Rs. 18,000 per month, depending on the marks obtained and the type of scholarship exam (SAKSHAM, SWABHIMAN, SAMADHAN, etc.).

Since you already have a 12th grade qualification with 84%, you meet the qualification criteria and are eligible to apply for the Medhavi Scholarship exam. Make sure to prepare well for the exam to maximize your chances of receiving a higher scholarship.

Hope you find this useful!

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Yuvan 30 August,2024

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

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Central Board of Secondary Education Class 12th Examination

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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

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

NCERT Solutions For Class 12 Maths Chapter 10 Exercise 10.3

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