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Test: Surface Integral - Electrical Engineering (EE) MCQ


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10 Questions MCQ Test - Test: Surface Integral

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

Gauss law for electric field uses surface integral. State True/False 

Detailed Solution for Test: Surface Integral - Question 1

Answer: a
Explanation: Gauss law states that the electric flux passing through any closed surface is equal to the total charge enclosed by the surface. Thus the charge is defined as a surface integral.

Test: Surface Integral - Question 2

Surface integral is used to compute

Detailed Solution for Test: Surface Integral - Question 2

Answer: b
Explanation: Surface integral is used to compute area, which is the product of two quantities length and breadth. Thus it is two dimensional integral.

Test: Surface Integral - Question 3

 Coulomb’s law can be derived from Gauss law. State True/ False 

Detailed Solution for Test: Surface Integral - Question 3

Answer: a
Explanation: Gauss law, Q = ∫∫D.ds
By considering area of a sphere, ds = r2sin θ dθ dφ.
On integrating, we get Q = 4πr2D and D = εE, where E = F/Q.
Thus, we get Coulomb’s law F = Q1 x Q2/4∏εR2.

Test: Surface Integral - Question 4

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

Detailed Solution for Test: Surface Integral - Question 4

Answer: c
Explanation: ∫∫ ( 5r2/4) . (r2 sin θ dθ dφ), which is the integral to be evaluated.
Put r = 4m and substitute θ = 0→ π/4 and φ = 0→ 2π, the integral evaluates to 588.9.

Test: Surface Integral - Question 5

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

Detailed Solution for Test: Surface Integral - Question 5

Answer: d
Explanation: ∫∫ D.ds = ∫∫ (10ρ3/4).(ρ dφ dz), which is the integral to be evaluated. Put ρ = 4m, z = 0→5 and φ = 0→2π, the integral evaluates to 6400π.

Test: Surface Integral - Question 6

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

Detailed Solution for Test: Surface Integral - Question 6

Answer: c
Explanation: ∫∫ ( 5r2/4) . (r2 sin θ dθ dφ), which is the integral to be evaluated. Since it is double integral, we need to keep only two variables and one constant compulsorily. Evaluate it as two integrals keeping r = 1 for the first integral and r = 2 for the second integral, with φ = 0→2π and θ = 0→ π. The first integral value is 80π, whereas second integral gives -5π. On summing both integrals, we get 75π.

Test: Surface Integral - Question 7

Find the value of divergence theorem for A = xy2 i + y3 j + y2z k for a cuboid given by 0<x<1, 0<y<1 and 0<z<1.

Detailed Solution for Test: Surface Integral - Question 7

Answer: c
Explanation: A cuboid has six faces. ∫∫A.ds = ∫∫Ax=0 dy dz + ∫∫Ax=1 dy dz + ∫∫Ay=0 dx dz + ∫∫Ay=1 dx dz + ∫∫Az=0 dy dx + ∫∫Az=1 dy dx. Substituting A and integrating we get (1/3) + 1 + (1/3) = 5/3.

Test: Surface Integral - Question 8

The ultimate result of the divergence theorem evaluates which one of the following?

Detailed Solution for Test: Surface Integral - Question 8

Answer: d
Explanation: Gauss law states that the electric flux passing through any closed surface is equal to the total charge enclosed by the surface. Thus, it is given by, ψ = ∫∫ D.ds= Q, where the divergence theorem computes the charge and flux, which are both the same.

Test: Surface Integral - Question 9

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: Surface Integral - Question 9

Answer: b
Explanation: While evaluating surface integral, there has to be two variables and one constant compulsorily. ∫∫D.ds = ∫∫Dx=0 dy dz + ∫∫Dx=1 dy dz + ∫∫Dy=0 dx dz + ∫∫Dy=2 dx dz + ∫∫Dz=0 dy dx + ∫∫Dz=3 dy dx. Put D in equation, the integral value we get is 12.

Test: Surface Integral - Question 10

 If D = 2xy i + 3yz j + 4xz k, how much flux passes through x = 3 plane for which -1<y<2 and 0<z<4?

Detailed Solution for Test: Surface Integral - Question 10

Answer: c
Explanation: By Gauss law, ψ = ∫∫ D.ds, where ds = dydz i at the x-plane. Put x = 3 and integrate at -1<y<2 and 0<z<4, we get 12 X 3 = 36.

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