Test: Volume Integral - Electrical Engineering (EE) MCQ

# Test: Volume Integral - Electrical Engineering (EE) MCQ

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## 10 Questions MCQ Test Electromagnetic Fields Theory (EMFT) - Test: Volume Integral

Test: Volume Integral for Electrical Engineering (EE) 2024 is part of Electromagnetic Fields Theory (EMFT) preparation. The Test: Volume Integral questions and answers have been prepared according to the Electrical Engineering (EE) exam syllabus.The Test: Volume Integral MCQs are made for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Volume Integral below.
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Test: Volume Integral - Question 1

### The divergence theorem converts

Detailed Solution for Test: Volume Integral - Question 1

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.

Test: Volume Integral - Question 2

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

Detailed Solution for Test: Volume Integral - Question 2

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.

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Test: Volume Integral - Question 3

### The volume integral is three dimensional. State True/False

Detailed Solution for Test: Volume Integral - Question 3

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

Test: Volume Integral - Question 4

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

Detailed Solution for Test: Volume Integral - Question 4

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).

Test: Volume Integral - Question 5

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

Detailed Solution for Test: Volume Integral - Question 5

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.

Test: Volume Integral - Question 6

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

Detailed Solution for Test: Volume Integral - Question 6

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.

Test: Volume Integral - 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.

Detailed Solution for Test: Volume Integral - Question 7

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.

Test: Volume Integral - Question 8

Using volume integral, which quantity can be calculated?

Detailed Solution for Test: Volume Integral - Question 8

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

Test: Volume Integral - 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.

Detailed Solution for Test: Volume Integral - Question 9

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.

Test: Volume Integral - 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.

Detailed Solution for Test: Volume Integral - Question 10

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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## Electromagnetic Fields Theory (EMFT)

11 videos|45 docs|73 tests