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# Vector Calculus - 1

## 20 Questions MCQ Test Topic-wise Tests & Solved Examples for IIT JAM Mathematics | Vector Calculus - 1

Description
This mock test of Vector Calculus - 1 for Mathematics helps you for every Mathematics entrance exam. This contains 20 Multiple Choice Questions for Mathematics Vector Calculus - 1 (mcq) to study with solutions a complete question bank. The solved questions answers in this Vector Calculus - 1 quiz give you a good mix of easy questions and tough questions. Mathematics students definitely take this Vector Calculus - 1 exercise for a better result in the exam. You can find other Vector Calculus - 1 extra questions, long questions & short questions for Mathematics on EduRev as well by searching above.
QUESTION: 1

### For a scalar function (x, y, z) = x2 + 3y2 + 2z2, the directional derivative at the point P( 1, 2, -1) is the direction of a vector  is

Solution:

We have,

So, the direction derivative in the direction of

QUESTION: 2

### Use the divergence theorem the value of  where, S is any closed surface enclosing volume V.

Solution:

where, is an outward drawn unit normal vector to S.

QUESTION: 3

### Find the value of

Solution:

So, from vector identity

QUESTION: 4

If   then the value of div  at the point (1, 1, -1) will be

Solution:

QUESTION: 5

Apply Stoke’s theorem, the value of  where C is the boundary of the triangle with vertices (2, 0, 0), (0, 3, 0) and (0, 0, 6) is

Solution:

Taking projection on three planes, we note that the surface S consists of three triangles, Δ OAB in XT- plane, Δ OBC in TZ-plane and Δ OAC in XZ-plane. Using two point formula, the equation of the line AB, BC, CA are respectively 3x + 2y = 6 , 2y + z = 6 , 3x + z = 6

So, by Stake’s theorem

QUESTION: 6

The line integral   from the origin to the point P( 1,1,1) is

Solution:

So, potential function of

So, line integral of the vector from point (0, 0, 0) to (1, 1, 1) is

QUESTION: 7

If  are to arbitrary vectors with magnitudes a and b respectively,  will be equal to

Solution:

Cross checking from option (a),

QUESTION: 8

Divergence of the three-dimensional radial vector field  is

Solution:

QUESTION: 9

Which of the following holds for any non-zero vector

Solution:

QUESTION: 10

A vector normal to  is

Solution:

We take,

= 1 - 2 + 1 = 0
So, B is normal to A.

QUESTION: 11

If   and curve C is the arc of the curve y = x3 from (0, 0 ) to (2, 8), then the value of

Solution:

Since, C is the curve y = x3 from (0, 0) to (2, 8)
So, let x = t ⇒ y = t3
If is the position vector of any point on C, then

or
or
At (0, 0) ⇒ t = x = 0 and at (2, 8) ⇒ t = 2

QUESTION: 12

The value of  by Stoke’s theorem, where   and C is the boundary of the triangle with vertices at ( 0 ,0 , 0 ) , ( 1 , 0 , 0 ) and ( 1 ,1 , 0 ) is

Solution:

We have, curl

Also we note that z coordinate of each vertex of the triangle is zero.
or The triangle lies in the xy-plane. So,
So, curl
In the figure, we have only considered the xy-plane. So, by Stoke’s Theorem

QUESTION: 13

If  is the reciprocal system to the vectors   then the value of   is

Solution:

Since,

Therefore,

QUESTION: 14

Scalar triple product  is equal to

Solution:

QUESTION: 15

R is a closed planar region as shown by the shaded area in the figure below. Its boundary C consists of the circles C1 and C2.

If  are all continuous everywhere in R, Green’s theorem states that

Which one of the following alternatives correctly depicts the direction of integration along C?

Solution:

The region R is bounded by two closed circles C1 and C2, so it is doubly connected. To apply it in Green’s theorem, we need to convert it into simply connected region. For it, we apply cut AD and consider the region R having simple closed curve ABCADEFDA in the anticlockwise direction. So, the directions shown in figure, (c) is correct option.

QUESTION: 16

The value of  is

Solution:

Since,
Putting

QUESTION: 17

The vector field  are unit vectors) is

Solution:

Given,
Now, for divergence

Hence, vector field is divergence-free.
Now, for irrotational

QUESTION: 18

Use Gauss’s divergence theorem to find  where  and S is the closed surface in the first octant bounded by y2 + z2 = 9 and x = 2.

Solution:

Let V be the volume enclosed by the closed surface S, i.e., the volume in the first octant bounded by the cylinder y2 I z2 = 9 and the planes x = 0, x = 2. Then by Gauss’s divergence theorem, we have

QUESTION: 19

For the scalar field  magnitude of the gradient at the point (1, 3) is

Solution:

Since,
So,
At (1, 3),

So,

QUESTION: 20

is equal to

Solution: