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# Vector Calculus MCQ Level - 1

## 10 Questions MCQ Test Topic wise Tests for IIT JAM Physics | Vector Calculus MCQ Level - 1

Description
This mock test of Vector Calculus MCQ Level - 1 for IIT JAM helps you for every IIT JAM entrance exam. This contains 10 Multiple Choice Questions for IIT JAM Vector Calculus MCQ Level - 1 (mcq) to study with solutions a complete question bank. The solved questions answers in this Vector Calculus MCQ Level - 1 quiz give you a good mix of easy questions and tough questions. IIT JAM students definitely take this Vector Calculus MCQ Level - 1 exercise for a better result in the exam. You can find other Vector Calculus MCQ Level - 1 extra questions, long questions & short questions for IIT JAM on EduRev as well by searching above.
QUESTION: 1

### The angle between the  x2 + y2 + z2 = 9  and  z = x2 + y2 – 3  at the point (2, –1, 2) is :

Solution:

The angle between the surface at point is the angle between the normal to the surface at the point.

A normal to x2 + y2 + z2 = 9 at (2, -1, 2) is A normal to z = x2 + y2 - 36 at (2, -1, 2) is We know that, is the required angle The correct answer is: QUESTION: 2

### For where C is the square in xy–plane projected from the cube x = 0, x = 2, y = 0, y = 2, z = 0, z = 2 above xy–plane, will be equal to :

Solution:

By Stoke’s theorem,  ∴  we need to evaluate   = –4

QUESTION: 3

### If and C is the curve  y = x3  from the point (1, 1) to (2, 8), then will be :

Solution:  QUESTION: 4

The value of where and S in the surface of the plane 2x + y + 2z = 6 in the first octant will be

Solution: Normal to the surface = constant will be :    QUESTION: 5

A vector field which has a vanishing divergence is called as ____________

Solution:

By the definition: A vector field whose divergence comes out to be zero or Vanishes is called as a Solenoidal Vector Field. i.e.

If is a Solenoidal Vector field.

QUESTION: 6

The value of the line integral where, C is the boundary of the region lying between the squares with vertices (1, 1), (–1, 1), (–1, –1) and (1, –1) and (2, 2), (–2, 2), (–2, –2) and (2, -2) will be :

Solution:

By Green’s Theorem, Explanation : ∫3x2ey dx + ey dy

= ∫∫-3x2ey dxdy

= -3*4 ∫(1 to 2)x2 dx ∫(1 to 2)ey dy

= 12[x3/3](1 to 2) [ey](1 to 2)

= -4(8-1)(e2-e)

= -28(e2-e)

QUESTION: 7

The value of where, and S is the surface of the parallelepiped bounded by x = 0, y = 0, z = 0, x = 2, y = 1, z = 3 will be :

Solution: By Gauss Divergence Theorem,   QUESTION: 8

If and then (a, b) =

Solution:   which is given to be Hence, for b = 2 and a being any value.

The correct answer is: (1, 2)

QUESTION: 9 is equal to :

Solution:  The correct answer is: QUESTION: 10

The value of where C is the intersection of  z = x + 4 with x2 + y2 = 4  will be :

Solution:  Also, the normal to the surface z – x = constant is   