Test: Green’s Theorem


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10 Questions MCQ Test Electromagnetic Fields Theory (EMFT) | Test: Green’s Theorem

Test: Green’s Theorem for Electrical Engineering (EE) 2023 is part of Electromagnetic Fields Theory (EMFT) preparation. The Test: Green’s Theorem questions and answers have been prepared according to the Electrical Engineering (EE) exam syllabus.The Test: Green’s Theorem MCQs are made for Electrical Engineering (EE) 2023 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Green’s Theorem below.
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Test: Green’s Theorem - Question 1

Mathematically, the functions in Green’s theorem will be

Detailed Solution for Test: Green’s Theorem - Question 1

Answer: c
Explanation: The Green’s theorem states that if L and M are functions of (x,y) in an open region containing D and having continuous partial derivatives then,
∫ (F dx + G dy) = ∫∫(dG/dx – dF/dy)dx dy, with path taken anticlockwise.

Test: Green’s Theorem - Question 2

Find the value of Green’s theorem for F = x2 and G = y2 is

Detailed Solution for Test: Green’s Theorem - Question 2

Answer: a
Explanation: ∫∫(dG/dx – dF/dy)dx dy = ∫∫(0 – 0)dx dy = 0. The value of Green’s theorem gives zero for the functions given.

Test: Green’s Theorem - Question 3

Which of the following is not an application of Green’s theorem?

Detailed Solution for Test: Green’s Theorem - Question 3

Answer: c
Explanation: In physics, Green’s theorem is used to find the two dimensional flow integrals. In plane geometry, it is used to find the area and centroid of plane figures.

Test: Green’s Theorem - Question 4

The path traversal in calculating the Green’s theorem is

Detailed Solution for Test: Green’s Theorem - Question 4

Answer: b
Explanation: The Green’s theorem calculates the area traversed by the functions in the region in the anticlockwise direction. This converts the line integral to surface integral.

Test: Green’s Theorem - Question 5

Calculate the Green’s value for the functions F = y2 and G = x2 for the region x = 1 and y = 2 from origin.

Detailed Solution for Test: Green’s Theorem - Question 5

Answer: c
Explanation: ∫∫(dG/dx – dF/dy)dx dy = ∫∫(2x – 2y)dx dy. On integrating for x = 0->1 and y = 0->2, we get Green’s value as -2.

Test: Green’s Theorem - Question 6

If two functions A and B are discrete, their Green’s value for a region of circle of radius a in the positive quadrant is

Detailed Solution for Test: Green’s Theorem - Question 6

Answer: d
Explanation: Green’s theorem is valid only for continuous functions. Since the given functions are discrete, the theorem is invalid or does not exist.

Test: Green’s Theorem - Question 7

Applications of Green’s theorem are meant to be in

Detailed Solution for Test: Green’s Theorem - Question 7

Answer: b
Explanation: Since Green’s theorem converts line integral to surface integral, we get the value as two dimensional. In other words the functions are variable with respect to x,y, which is two dimensional.

Test: Green’s Theorem - Question 8

The Green’s theorem can be related to which of the following theorems mathematically?

Detailed Solution for Test: Green’s Theorem - Question 8

Answer: b
Explanation: The Green’s theorem is a special case of the Kelvin- Stokes theorem, when applied to a region in the x-y plane. It is a widely used theorem in mathematics and physics.

Test: Green’s Theorem - Question 9

he Shoelace formula is a shortcut for the Green’s theorem. State True/False. 

Detailed Solution for Test: Green’s Theorem - Question 9

Answer: a
Explanation: The Shoelace theorem is used to find the area of polygon using cross multiples. This can be verified by dividing the polygon into triangles. It is a special case of Green’s theorem.

Test: Green’s Theorem - Question 10

Find the area of a right angled triangle with sides of 90 degree unit and the functions described by L = cos y and M = sin x.

Detailed Solution for Test: Green’s Theorem - Question 10

Answer: d
Explanation: dM/dx = cos x and dL/dy = -sin y
∫∫(dM/dx – dL/dy)dx dy = ∫∫ (cos x + sin y)dx dy. On integrating with x = 0->90 and y = 0->90, we get area of right angled triangle as -180 units (taken in clockwise direction). Since area cannot be negative, we take 180 units.

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