Vector Calculus - 2

# Vector Calculus - 2

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## 20 Questions MCQ Test Topic-wise Tests & Solved Examples for IIT JAM Mathematics | Vector Calculus - 2

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Vector Calculus - 2 - Question 1
Detailed Solution for Vector Calculus - 2 - Question 1

Vector Calculus - 2 - Question 2

### then the value of  where C is the curve in the XY-plane, y = 2x2 from (0,0) to (1,2) is

Detailed Solution for Vector Calculus - 2 - Question 2

Explanation : y = 2x2

dy/dx = 4x

So F. dr

∫(x = 0 to 1) [3x(2x2)dx . (2x2)2 d(2x2)]

∫(x = 0 to 1) (6x2 - 16x5)dx

= -7/6

Vector Calculus - 2 - Question 3

### Value of the integral  where C is the square cut from the first quadrant by the lines x = 1 and y = 1 will be (use Green’s theorem to change the line integral into double integral)

Detailed Solution for Vector Calculus - 2 - Question 3

We know that in Green’s theorem,

On comparing, we get M = -y2 and N = xy

Vector Calculus - 2 - Question 4

The divergence of the vector field

Detailed Solution for Vector Calculus - 2 - Question 4

Vector Calculus - 2 - Question 5

The directional derivative of f(x, y, z) = x2 + y2 + z2 at the point (1, 1, 1) in the direction

Detailed Solution for Vector Calculus - 2 - Question 5

We have,

Thus, directional derivative of f in the direction of   the point P(1, 1, 1),

Vector Calculus - 2 - Question 6

The unit normal vector to the surface of the sphere x2 + y2 + z2 = 1 at the point  and  are unit normal vectors in the Cartesian coordinate system)

Detailed Solution for Vector Calculus - 2 - Question 6

Problem is to find unit normal vector to the surface of the sphere
f = x2 + y2 + z2 - 1 = 0
We know unit normal vector  is given by

Vector Calculus - 2 - Question 7

If  over the path shown in the figure is

Detailed Solution for Vector Calculus - 2 - Question 7

Vector Calculus - 2 - Question 8

Apply Green’s theorem the value of  where C is the boundary of the area enclosed by the X-axis and the upper half of the circle x2 + y2 = a2 is

Detailed Solution for Vector Calculus - 2 - Question 8

Since, From Green’s theorem, we have

(A is the region of the figure)

On changing in polar coordinates, we get

Vector Calculus - 2 - Question 9

A fluid element has a velocity  The motion at (x, y) =

Detailed Solution for Vector Calculus - 2 - Question 9

or u = - y2x, v = 2yx2

In two-dim ensional flow, equation of continuity

Fluid is incompressible at this point.

Fluid flow is rotational.
Thus, fluid flow at  is rotational and incompressible.

Vector Calculus - 2 - Question 10

Unit vectors in X and Z-directions are  respectively. Which one of the following is the directional derivative of the function F(x, z) = In (x2 + z2) at the point P(4, 0), in the direction of

Detailed Solution for Vector Calculus - 2 - Question 10

Given, F(x, y) = In(x2 + z2) = log (y + z2)

Coordinate of point p is (4, 0).

or

Vector Calculus - 2 - Question 11

For a scalar function f(x, y, z) = x2 + 3y2 + 2z2, the gradient at the point P(1, 2, -1) is

Detailed Solution for Vector Calculus - 2 - Question 11

Vector Calculus - 2 - Question 12

then the value of div Curl is

Detailed Solution for Vector Calculus - 2 - Question 12

Vector Calculus - 2 - Question 13

The value of α for which the following three vectors are coplanar is

Detailed Solution for Vector Calculus - 2 - Question 13

Given,

Vectors  are coplanar, if

Vector Calculus - 2 - Question 14

Apply Green’ s theorem the value of where C is the square formed by the lines y = ±1, x = ±1 is

Detailed Solution for Vector Calculus - 2 - Question 14

Since, On comparing  with  we get M = x2 + xy and
N = x2 + y2
So, from Green’s theorem

Vector Calculus - 2 - Question 15

The divergence of the vector field  at a point (1, 1, 1) is equal to

Detailed Solution for Vector Calculus - 2 - Question 15

Since, We have
On comparing Eq. (r) with

Vector Calculus - 2 - Question 16

is the reciprocal system to the vectors   then the value of

Vector Calculus - 2 - Question 17

where  are constant vectors then  is equal to

Vector Calculus - 2 - Question 18

, then the value of div curl   is

Detailed Solution for Vector Calculus - 2 - Question 18

Since div(curl⇀v)=0, the net rate of flow in vector field curl⇀v\) at any point is zero. Taking the curl of vector field ⇀F eliminates whatever divergence was present in ⇀F

Vector Calculus - 2 - Question 19

The value of div   will be

Detailed Solution for Vector Calculus - 2 - Question 19

If P (x,y,z) is a variable point in a three- dimensional space, O the origin, i, j and k the unit vectors along the x-axis, y-axis and z-axis respectively, then the vector OP given by x i + y j +z k is called the position vector of the point P and is denoted by r.

Divergence of any vector f = f1i+ f2j +f3k denoted by div f is defined as the scalar ( delta f1)/(delta x) + (delta f2)/(delta y) + (delta f3)/(delta z)

where the delta s denote partial derivatives.

Using this definition we find that div r = delta (x)/ delta x +delta (y)/delta y + delta (z)/ delta z = 1 + 1 +1 = 3.

Hence, divergence of a position vector = div r = 3.

Vector Calculus - 2 - Question 20

Divergence operators is defined for

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