Test: Volume Integral


10 Questions MCQ Test Electromagnetic Theory | Test: Volume Integral


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

The divergence theorem converts

Solution:

Answer: b
Explanation: The divergence theorem is given by, ∫∫ D.ds = ∫∫∫ Div (D) dv. It is clear that it converts surface (double) integral to volume(triple) integral.

QUESTION: 2

The triple integral is used to compute volume. State True/False 

Solution:

Answer: a
Explanation: The triple integral, as the name suggests integrates the function/quantity three times. This gives volume which is the product of three independent quantities.

QUESTION: 3

The volume integral is three dimensional. State True/False

Solution:

Answer: a
Explanation: Volume integral integrates the independent quantities by three times. Thus it is said to be three dimensional integral or triple integral.

QUESTION: 4

Find the charged enclosed by a sphere of charge density ρ and radius a. 

Solution:

Answer: b
Explanation: The charge enclosed by the sphere is Q = ∫∫∫ ρ dv.
Where, dv = r2 sin θ dr dθ dφ and on integrating with r = 0->a, φ = 0->2π and θ = 0->π, we get Q = ρ(4πa3/3).

QUESTION: 5

Evaluate Gauss law for D = 5r2/2 i in spherical coordinates with r = 4m and θ = π/2 as volume integral.

Solution:

Answer: b
Explanation: ∫∫ D.ds = ∫∫∫ Div (D) dv, where RHS needs to be computed.
The divergence of D given is, Div(D) = 5r and dv = r2 sin θ dr dθ dφ. On integrating, r = 0->4, φ = 0->2π and θ = 0->π/4, we get Q = 588.9.

QUESTION: 6

Compute divergence theorem for D = 5r2/4 i in spherical coordinates between r = 1 and r = 2 in volume integral.

Solution:

Answer: c
Explanation: D.ds = ∫∫∫ Div (D) dv, where RHS needs to be computed.
The divergence of D given is, Div(D) = 5r and dv = r2 sin θ dr dθ dφ. On integrating, r = 1->2, φ = 0->2π and θ = 0->π, we get Q = 75 π.

QUESTION: 7

Compute the Gauss law for D = 10ρ3/4 i, in cylindrical coordinates with ρ = 4m, z = 0 and z = 5, hence find charge using volume integral.

Solution:

Answer: d
Explanation: Q = D.ds = ∫∫∫ Div (D) dv, where RHS needs to be computed.
The divergence of D given is, Div(D) = 10 ρ2 and dv = ρ dρ dφ dz. On integrating, ρ = 0->4, φ = 0->2π and z = 0->5, we get Q = 6400 π.

QUESTION: 8

Using volume integral, which quantity can be calculated?

Solution:

Answer: c
Explanation: The volume integral gives the volume of a vector in a region. Thus volume of a cube can be computed.

QUESTION: 9

Compute the charge enclosed by a cube of 2m each edge centered at the origin and with the edges parallel to the axes. Given D = 10y3/3 j.

Solution:

Answer: c
Explanation: Div(D) = 10y2
∫∫∫Div (D) dv = ∫∫∫ 10y2 dx dy dz. On integrating, x = -1->1, y = -1->1 and z = -1->1, we get Q = 80/3.

QUESTION: 10

Find the value of divergence theorem for the field D = 2xy i + x2 j for the rectangular parallelepiped given by x = 0 and 1, y = 0 and 2, z = 0 and 3.

Solution:

Answer: b
Explanation: Div (D) = 2y
∫∫∫Div (D) dv = ∫∫∫ 2y dx dy dz. On integrating, x = 0->1, y = 0->2 and z = 0->3, we get Q = 12.

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