RD Sharma Class 12 Exercise 22.5 Algebra of vectors Solutions Maths - Download PDF Free Online

# RD Sharma Class 12 Exercise 22.5 Algebra of vectors Solutions Maths - Download PDF Free Online

Edited By Kuldeep Maurya | Updated on Jan 24, 2022 06:19 PM IST

RD Sharma's book has set a standard for other publications as well. Chapter 22 of this book is Algebra of Vectors which includes a detailed study of vector quantities. Vector quantities are those that have magnitude as well as direction as compared to scalar amounts which have the only magnitude. This book could be dicey for some students, which is why RD Sharma Class 12th Exercise 22.5 has arrived to help.

RD Sharma Class 12th Chapter 22 Exercise 22.5 has exercise-wise questions in this book. This particular exercise has 16 questions from the chapter 'Algebra of Vectors' along with the examples. The concepts that are discussed are position vectors and the components of vectors in three dimensions.

## Algebra of Vectors Excercise:22.5

Algebra of vector exercise 22.5 question 1

5
Hints: $\left | x\hat{i} +y\hat{j}\right |=\sqrt{x^{2}+y^{2}}=\left | \vec{a} \right |$
Given: Position vector of a point (-4,-3) be $\left | \vec{a} \right |$
Find: $\left | \vec{a} \right |$
We know position vector of a point(x,y) is given by $x\hat{i} +y\hat{j}$ where $\hat{i}$ and $\hat{j}$ are units vectors in x and y directions.
$\vec{a}=(-4)\hat{i} +(-3)\hat{j}$
Now we need to find magnitude of $\vec{a}$ i.e $\left |\vec{a} \right |$
$\left | x\hat{i} +y\hat{j}\right |=\sqrt{x^{2}+y^{2}}$
Here, x=-4 and y=-3
$\left |\vec{a} \right |=\sqrt{(-4)^{2} +(-3)^{2}}\\ \left |\vec{a} \right |=\sqrt{16+9}\\ \left |\vec{a} \right |=\sqrt{25}\\ \left |\vec{a} \right |=5$
thus $\left |\vec{a} \right |=5$

Algebra of vector exercise 22.5 question 2

$n=\pm 5$
Hints: $\left | x\hat{i} +y\hat{j}\right |=\sqrt{x^{2}+y^{2}}=\left | \vec{a} \right |$
Given: Position vector of $\vec{a}$ point (12,n) be $\left | \vec{a} \right |$
Find: $\left | \vec{a} \right |$
We know position vector of a point(x,y) is given by $x\hat{i} +y\hat{j}$ where $\hat{i}$ and $\hat{j}$ are units vectors in x and y directions.
$\vec{a}=12\hat{i} +n\hat{j}$
Then:
$\left | \vec{a} \right |=\sqrt{x^{2}+y^{2}}$
Here, x=12 and y=n
$\left |\vec{a} \right |=\sqrt{12^{2} +n^{2}}\\ \left |\vec{a} \right |=13$[given]
$13=\sqrt{(12)^{2} +(n)^{2}}\\$
Squaring both sides
$13^{2}=(12)^{2} +(n)^{2}\\ 169=144+n^{2}\\ n^{2}=25\\ n=\pm 5$

Algebra of vector exercise 22.5 question 4 (i)

$\vec{AB}=-3\hat{i}+4\hat{j}\\ \left |\vec{AB} \right |=5$
Hint: $\left | x\hat{i} +y\hat{j}\right |=\sqrt{x^{2}+y^{2}}=\left | \vec{a} \right |$
Position vector $\left |\vec{AB} \right |$ is given x, y
$\left |\vec{AB} \right |$ = position vector of B - position vector of A
Given:
A=(4,-1) and B(1,3)
Solution:
We know position vector of a point (x, y) is given by $x\hat{i} +y\hat{j}$ where $\hat{i}$ and $\hat{j}$ are unit vector in
x and y direction
$A=4\hat{i}+(-1)\hat{j}\\ B=i+3\hat{j}$
Then the position vector $\vec{AB}$ is given by
$\vec{AB}$ = position vector of B - position vector of A
$=\left (\hat{i}+3\hat{j} \right )-\left (4\hat{i}-\hat{j} \right )\\ =\hat{i}+3\hat{j} -4\hat{i}+\hat{j} \\ AB=-3\hat{i}+4j$
So,
$\left |\vec{AB} \right |=(-3)^{2}+(4)^{2}\\ =\sqrt{9+16}=\sqrt{25}\\\left |\vec{AB} \right |=5$

Algebra of vector exercise 22.5 question 4 (ii)

$\vec{AB}=4\hat{i}-8\hat{j}\\ \left |\vec{AB} \right |=4\sqrt{5}$
Hint: $\left | x\hat{i} +y\hat{j}\right |=\sqrt{x^{2}+y^{2}}=\left | \vec{a} \right |$
Position vector $\vec{AB}$ is given x, y
$\vec{AB}$ = position vector of B - position vector of A
Given:
A=(-6,3) and B(-2,-5)
Solution:
We know position vector of a point (x, y) is given by $x\hat{i} +y\hat{j}$ where $\hat{i}$ and $\hat{j}$ are unit vector in
x and y direction
$\vec{A}=-6\hat{i}+3\hat{j}\\ \vec{B}=-2i-5\hat{j}$
Then the position vector $\vec{AB}$ is given by
$\vec{AB}$ = position vector of B - position vector of A
$=\left (2\hat{i}-5\hat{j} \right )-\left (-6\hat{i}+3\hat{j} \right )\\ =-2\hat{i}-5\hat{j} +6\hat{i}-3\hat{j} \\ =4\hat{i}-8\hat{j}\\ \left |\vec{AB} \right |=\sqrt{4^{2}+(-8)^{2}}=\sqrt{16+64}\\ =\sqrt{80}\\ \left |\vec{AB} \right |=4\sqrt{5}$-
(-1,-2)
Hint: Position vector $\vec{AB}$ is given $\vec{AB}$ = position vector B- position vector of A
Given: Coordinate of A (-1, 3)
Coordinates of B (-2,1)
Solution:
$\vec{A}= -\hat{i}+3\hat{j}$ & $\vec{B}= -2\hat{i}+\hat{j}$
We know position vector of point (x, y) is given by $(x\hat{i}+y\hat{j})$
Let 0 be the origin
Let P(x,y) be the second point
Then $\vec{P}$ is the tip of the position vector $\vec{OP}$ of the point p
We have: $\vec{OP}=x\hat{i}+y\hat{j}$
position vector of B- position vector of A
$\vec{AB}$= position vector of B= position vector of A
$=(-2\hat{i}+\hat{j})-(-\hat{i}+3\hat{j})\\ =-2\hat{i}+\hat{j}+\hat{i}-3\hat{j}\\ =-\hat{i}-2\hat{j}$
$\vec{OP}=\vec{AB}$ given
$x\hat{i}+y\hat{j}=-\hat{i}-2\hat{j}$
then x=-1 and y= -2
Hence coordinate of the required point is (-1,-2)
the coordinates of D is (-4,-3)
Hint: Position vectors $\vec{AB}$ is given by = position vector of B -position vector of A
Given;
Coordinates of A=(-2,-1)
Coordinates of B=(3,0)
Coordinates of C=(1,-2)
Let the coordinates of D is (x , y)
Since ABCD is a parallelogram,
AB=DC
We have
$\vec{AB}=\vec{DC}$
As we know position vector of point (x, y) is given by $x\hat{i}=y\hat{j}$where $\hat{i}$ and $\hat{j}$ are position vector of x, and y direction.
$\vec{A}=-2\hat{i}-\hat{j}\\ \vec{B}=3\hat{i}+0\hat{j}\\ \vec{C}=\hat{i}-2\hat{j}\\ \vec{D}=x\hat{i}+y\hat{j}$
Then the position vector $\vec{AB}$ is given by
$\vec{AB}$= position vector of B - position vector of A
$\left (3\hat{i}+0\hat{j} \right )-\left (-2\hat{i}-\hat{j} \right )\\ 3\hat{i}+0\hat{j} +2\hat{i}+\hat{j} \\ \vec{AB}=5\hat{i}+\hat{j}$
Then we find $\vec{DC}$
$\vec{DC}$= position vector C- position vector of D
$=\left (\hat{i}-2\hat{j} \right )-\left (x\hat{i}+y\hat{j} \right )\\ =\hat{i}-x\hat{i}-2\hat{j}+y\hat{j} \\ =\hat{i}\left ( 1-x \right )+\hat{j}\left ( -2-y \right )$
We have $\vec{AB}=\vec{DC}$
Then:
$5\hat{i}+\hat{j}=\left ( 1-x \right )\hat{i}+\left ( -2-y \right )\hat{j}$
Then comparing coefficient of $\hat{i}$ and $\hat{j}$
$\begin{matrix} 5=1-x & 1=-2-y\\ x=1-5 &y=-2-1 \\ x=-4 & y=-3 \end{matrix}$
Hence the coordinates of D is (-4, -3)
$\hat{i}-5\hat{j}$
Hint: position vector of point (x , y) is given by where $\hat{i}$ and $\hat{j}$ are the position in x an y direction.
Given:
Point $A(3,4),B(5,-6),C(4,-1)$
Find $\vec{a}+2\vec{b}-3\vec{c}$
Solution:
As we know position vector of point (x, y) is given by $\left (x\hat{i}+y\hat{j} \right )$
Where $\hat{i}$and $\hat{j}$ are the position vector in x and y direction
Then,
$\vec{A}=\left (3\hat{i}+4\hat{j} \right )\\ \vec{B}=\left (5\hat{i}+(-6)\hat{j} \right )=5\hat{i}-6\hat{j}\\ \vec{C}=\left (4\hat{i}+(-1)\hat{j} \right )=4\hat{i}-\hat{j}\\ 2\vec{B}=2\left (5\hat{i}-6\hat{j} \right )=10\hat{i}-12\hat{j}\\ 3\vec{C}=3\left (4\hat{i}-\hat{j} \right )=12\hat{i}-3\hat{j}$
$\vec{a}+2\vec{b}-3\vec{c}$ Substitute value of $\vec{a},2\vec{b},3\vec{c}$ in equation $\vec{a}+2\vec{b}-3\vec{c}$
$=3\hat{i}+4\hat{j}+10\hat{i}-12\hat{j}-\left ( 12\hat{i}-3\hat{j} \right )\\ =3\hat{i}+4\hat{j}+10\hat{i}-12\hat{j}- 12\hat{i}+3\hat{j}\\ =\hat{i}\left ( 3+10-12 \right )+\hat{j}\left ( 4-12+3 \right )\\ =\hat{i}\left ( 1 \right )+\hat{j}\left ( -5 \right )\\ =\hat{i}-5\hat{j}$
(9, -4)
Hint: position vector $\vec{AB}$ is given by $\vec{AB}$ =position vector of B- position vector of A
Given: coordinates of A(4, -1) , a $\vec{a}$ is the position vector whose tip i is (5, -3)
Solution:
Let O be the origin of point (0, 0)
Position P(5, -3) be the tip of the position vector $\vec{a}$
Then the $\vec{a}$ ,$\vec{OP}$ =Position vector of p- position vector of o.
As we know position vector of point (x, y) is given by $\left ( x\hat{i} + y\hat{j}\right )$ where $\hat{i}$ and $\hat{j}$ are the unit vectors.
Then
$\vec{P}=5\hat{i}+-3\hat{j}=5\hat{i}-3\hat{j}\\ \vec{O}=0\hat{i}+0\hat{j} \\ \vec{OP}=5\hat{i}-3\hat{j}-0\hat{i}+0\hat{j}\\ \vec{OP}=5\hat{i}-3\hat{j}-0\hat{i}-0\hat{j}\\ =5-0\hat{i}+-3-0\hat{j}\\ \vec{OP}=5\hat{i}-3\hat{j}$
Let the coordinate of B be (x,y) and A has coordinate (4,-1)
Therefore
$\vec{A}=4\hat{i}+-1\hat{j}=4\hat{i}-\hat{j}\\ \vec{B}=x\hat{i}+y\hat{j}$
AB=$\vec{AB}$ Position vectors of B – position vector of A
$\vec{AB}=x\hat{i}+y\hat{j}-\left (4\hat{i}-\hat{j} \right )\\ =x\hat{i}+y\hat{j}-4\hat{i}+\hat{j}\\ =\left (x-4 \right )\hat{i}+\hat{j}\left (y+1 \right )$
Now,
$\vec{AB}=\vec{a}\\ \left ( x-4 \right )\hat{i}+\left ( y+1 \right )\hat{j}=5\hat{i}-3\hat{j}$
Comparing coefficient of $\hat{i}$ & $\hat{j}$
$x-4=5\\ x=5+4\\ x=9\\ x+1=-3\\ y=-3-1\\ y=-4$
Hence, the coordinate of B are (9,-4)
the magnitude of AB and AC is equal
Hence the points $2\hat{i},-\hat{i}-4\hat{j}$ and $-\hat{i}+4\hat{j}$ form an isosceles triangle
Hint: the magnitude of $\left |\vec{a} \right |=\sqrt{x^{2}+y^{2}}$
Position vector $\vec{AB}$ is given by the $\vec{AB}$ = position vector of B - position vector of A
Given: the points A,B,C with position vectors $\vec{a},\vec{b},\vec{c}$ respectively
Also,
$\vec{a}=2\hat{i}\\ \vec{b}=-\hat{i}-4\hat{j}\\ \vec{c}=-\hat{i}+4\hat{j}$
Solution:
Now we find the position vector $\vec{AB}$
Then,
$\vec{AB}=\vec{b}-\vec{a}\\ =\left (-\hat{i}-4\hat{j} \right )-2\hat{i}\\ =-\hat{i}-4\hat{j}-2\hat{i}\\ \vec{AB}=-3\hat{3}-4\hat{j}$
Now
$\left |\vec{AB} \right |=\sqrt{x^{2}+y^{2}}$
Here x=-3 and y= -4
$=\sqrt{(-3)^{2}+(-4)^{2}}\\ =\sqrt{9+16}=\sqrt{25}\\ =5$
$\vec{BC}$=Position vector of C – Position vector of B
$=\left ( -\hat{i}+4\hat{j} \right )-\left ( -\hat{i}-4\hat{j} \right )\\ = -\hat{i}+4\hat{j}+\hat{i}+4\hat{j} \\ =0\hat{i}+8\hat{j}\\ =8\hat{j}$
$\vec{AC}$=Position vector of C – position vector of A
$=\left ( -\hat{i}+4\hat{j} \right )-\left ( 2\hat{i} \right )\\ = -\hat{i}+4\hat{j}-\hat{j} \\ =-3\hat{i}+4\hat{j}\\$
Now
$\left |\vec{AC} \right |=\sqrt{x^{2}+y^{2}}$
Here x=-3 and y=4
$=\sqrt{(-3)^{2}+(-4)^{2}}\\ =\sqrt{9+16}=\sqrt{25}\\ =5$
Since, the magnitude of AB and AC are equal
Hence, the points $2\hat{i},-\hat{i}-4\hat{j}$ and $-\hat{i}+4\hat{j}$ form an isosceles triangle

Algebra of Vector exercise 22.5 question 10

Answer: $\frac{1}{2}\hat{i}+\frac{\sqrt{3}}{2}\hat{j}$
Hint: a unit vector parallel to
$\vec{a}=\hat{a}=\frac{\vec{a}}{\left | \vec{a} \right |}\\ \left | \vec{a} \right |=\sqrt{x^{2}+y^{2}}$
Given: $\vec{a}=\hat{i}+\sqrt{3}\hat{j}$
Solution:
We can find the magnitude of $\vec{a}$
$\left | \vec{a} \right |=\sqrt{x^{2}+y^{2}}$
Here x = 1 and $y=\sqrt{3}$
$\left |\vec{a} \right |=1^{2}+(\sqrt{3})^{2}\\ =\sqrt{1+3}=4\\ \left |\vec{a} \right |=2$
Unit vector parallel to $\vec{a}=\hat{a}=\frac{\vec{a}}{\left | \vec{a} \right |}\\$
$=\frac{1}{2}\left ( \hat{i}+\sqrt{3}\hat{j} \right )\\ =\frac{1}{2}\hat{i}+\frac{\sqrt{3}}{2}\hat{j}$

$\lambda=8$ and $\mu =-5$
Hint: here we use sector formula
$\frac{\left ( mx_{2}+nx_{1} \right )}{m+n},\frac{\left ( my_{2}+ny_{1} \right )}{m+n}$
Where m,n are the given ratios
Given: the position vectors of points A, B and C are
$\vec{A}=\lambda \hat{i}+3\hat{j}\\\vec{B}=12\hat{i}+\mu \hat{j}\\\vec{C}=11\hat{i}-3\hat{j}$
Solution:
It is given that C, divides the line segment joining A and B in the ratio 3:1
\begin{aligned} &11 \hat{\imath}-3 \hat{\jmath}=3 \times(12 \hat{\imath}+\mu \hat{\jmath})+1(\lambda \hat{\imath}+3 \hat{\jmath}) \\ &11 \hat{\imath}-3 \hat{\jmath}=\frac{36 \hat{\imath}+3 u \hat{\jmath}+\lambda \hat{\imath}+3 \hat{\jmath}}{4} \\ \end{aligned}
\begin{aligned}&11 \hat{\imath}-3 \hat{\jmath}=\frac{(36+\lambda) \hat{\imath}+(3 \mu+3) \hat{\jmath}}{4} \\ &4(11 \hat{\imath}-3 \hat{\jmath})=(36+\lambda) \hat{\imath}+(3 u+3) \hat{\jmath} \\ &44 \hat{\imath}-12 \hat{\jmath}=(36+\lambda) \hat{\imath}+(3 u+3) \hat{\jmath} \end{aligned}
Equating the corresponding components, we get;
$36+\lambda=44\\ \lambda=44-36\\ \lambda=8$
and
$3\mu+3=-12\\ 3\mu=-12-3=-15\\ \mu=\frac{-15}{3}=-5\\ \mu=-5$
Thus, the values of $\lambda$ and $\mu$ are 8 and -5 respectively
: The components of $\vec{P}$ along x axis is a vector of magnitude 3 along the positive direction of x axis
The component of $\vec{P}$ along y axis is a vector of magnitude 2 along the positive direction of Y axis
$3\hat{i},2\hat{j}$
Hint: position vector of point (x, y) is given by $x\hat{i}+y\hat{j}$ where $\hat{i}$ and $\hat{j}$ are the unit vector in x and y direction
Given: P(3,2)
Solution:
As we know, position vector of a point (x, y) is given by $x\hat{i}+y\hat{j}$
Then,
Position vector of a point (3, 2) is $\vec{p}=3\hat{i}+2\hat{j}$
The components of $\vec{P}$along x axis is a vector of magnitude 3 along the positive direction of x axis
The component of $\vec{P}$ along y axis is a vector of magnitude 2 along the positive direction of Y axis
The components of $\vec{Q}$ along x axis is a vector of magnitude 5 along the negative direction of x axis
The component of $\vec{Q}$ along y axis is a vector of magnitude 1 along the positive direction of Y axis
$-5\hat{i},\hat{j}$
Hint: position vector of point (x, y) is given by $x\hat{i}+y\hat{j}$ where $\hat{i}$ and $\hat{j}$ are the unit vector in x and y direction
Given:Q=(-5,1)
Solution:
As we know, position vector of a point (x, y) is given by $x\hat{i}+y\hat{j}$
So
$\vec{Q}=-5\hat{i}+1\hat{j}\\ \vec{Q}=-5\hat{i}+\hat{j}\\$
The components of $\vec{Q}$ along x axis is a vector of magnitude 5 along the negative direction of x axis
The component of $\vec{Q}$ along y axis is a vector of magnitude 1 along the positive direction of Y axis
The components of $\vec{R}$ along x axis is a vector of magnitude 11 along the negative direction of x axis
The component of $\vec{R}$ along y axis is a vector of magnitude 9 along the negative direction of Y axis
$-11\hat{i},-9\hat{j}$
Hint: position vector of point (x, y) is given by $x\hat{i}+y\hat{j}$ where $\hat{i}$ and $\hat{j}$ are the unit vector in x and y direction
Given:R=(-11,-9)
Solution:
As we know, position vector of a point (x, y) is given by $x\hat{i}+y\hat{j}$
So
$\vec{R}=-11\hat{i}+(-9)\hat{j}\\ \vec{R}=-11\hat{i}-9\hat{j}\\$
The components of $\vec{R}$ along x-axis is a vector of magnitude 11 along the negative direction of x axis
The component of $\vec{R}$ along y-axis is a vector of magnitude 9 along the negative direction of Y axis
The components of $\vec{S}$ along x axis is a vector of magnitude 4 along the positive direction of x axis
The component of $\vec{S}$ along y axis is a vector of magnitude 3 along the negative direction of Y axis
$4\hat{i},-3\hat{j}$
Hint: position vector of point (x, y) is given by $x\hat{i}+y\hat{j}$ where $\hat{i}$ and $\hat{j}$ are the unit vector in x and y direction
Given:S=(-4,-3)
Solution:
As we know, position vector of a point (x, y) is given by $x\hat{i}+y\hat{j}$
So
$\vec{S}=4\hat{i}+(-3)\hat{j}\\ \vec{S}=4\hat{i}-3\hat{j}\\$
The components of $\vec{S}$ along x-axis is a vector of magnitude 4 along the positive direction of x-axis
The component of $\vec{S}$ along y-axis is a vector of magnitude 3 along the negative direction of Y-axis

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## RD Sharma Chapter wise Solutions

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1. How many questions are there in each exercise of RD Sharma Solutions for Class 12 Maths Chapter 22?

The number of questions in each exercise of RD Sharma Solutions for Class 12 Maths Chapter 22 ais

Exercise 22.1 - 24 questions

Exercise 22.2 - 13 questions

Exercise 22.3 - 7 questions

Exercise 22.4 - 6 questions

Exercise 22.5 - 16 questions

Exercise 22.6 - 20 questions

Exercise 22.7 - 14 questions

Exercise 22.8 - 18 questions

Exercise 22.9 - 16 questions

MCQ - 27 questions

VSA - 52 questions

FBQ - 23 questions

2. What is a cross product in the Algebra of vectors?

The multiplication sign(x) between two vectors denotes a cross product. In a three-dimensional system, it is a binary vector operation. If P and Q are two vectors independent of each other, the result of their cross product (P x Q) is perpendicular to both vectors and normal to the plane containing both vectors. It is symbolized by;

P x Q = |P| |Q| sin θ

3. What do you mean by scalar and vector quantities?

Scalar quantities are those quantities that have only magnitude with no direction whereas vector quantities have both magnitude and direction.

4. Where can I find RD Sharma solutions for Class 12 Mathematics online?

At Career360, all the solutions for RD Sharma are present and you can visit the website for downloading the PDF free of cost. The material is a reliable source for preparation.

5. What are the basic concepts of vectors?

Position vectors and Direction Cosines are the basic concepts of vectors. You can visit Career36the 0 for the latest updated solutions for RD Sharma.

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