Green's Theorem MCQ Quiz in मल्याळम - Objective Question with Answer for Green's Theorem - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Last updated on Apr 16, 2025

നേടുക Green's Theorem ഉത്തരങ്ങളും വിശദമായ പരിഹാരങ്ങളുമുള്ള മൾട്ടിപ്പിൾ ചോയ്സ് ചോദ്യങ്ങൾ (MCQ ക്വിസ്). ഇവ സൗജന്യമായി ഡൗൺലോഡ് ചെയ്യുക Green's Theorem MCQ ക്വിസ് പിഡിഎഫ്, ബാങ്കിംഗ്, എസ്എസ്‌സി, റെയിൽവേ, യുപിഎസ്‌സി, സ്റ്റേറ്റ് പിഎസ്‌സി തുടങ്ങിയ നിങ്ങളുടെ വരാനിരിക്കുന്ന പരീക്ഷകൾക്കായി തയ്യാറെടുക്കുക

Latest Green's Theorem MCQ Objective Questions

Top Green's Theorem MCQ Objective Questions

Green's Theorem Question 1:

Evaluate c[(2x2y2)dx+(x2+y2)dy], when C is the boundary of the area enclosed by the x – axis and the upper half of the circle x2 + y2 = a2

  1. 83a3
  2. 43a3
  3. 163a3
  4. 23a3

Answer (Detailed Solution Below)

Option 2 : 43a3

Green's Theorem Question 1 Detailed Solution

By Green’s theorem

c(ϕdx+ψdy)=E(ψxϕy)dxdy

c[(2x2y2)dx+(x2+y2)dy]=E[x(x2+y2)y(2x2y2)]dxdy

=2E(x+y)dxdy

GATE MATHS LT 1 1

Now changing to polar co-ordinates, r varies from 0 to a and Q varies from 0 to π

=20a0πr(cosθ+sinθ)rdθdr=20ar2dr0π(cosθ+sinθ)dθ=20ar2dr.(2)=4r33]0a=4a33

Green's Theorem Question 2:

The area of the region enclosed by a simple closed curve C will be _________.

  1. 12c(1xdy1ydx)
  2. 12c(xdyydx)
  3. 12c(ydxxdy)
  4. 12c(dxx+dyy)

Answer (Detailed Solution Below)

Option 2 : 12c(xdyydx)

Green's Theorem Question 2 Detailed Solution

Concept:

Green's Theorem:

Green's Theorem relates a line integral around a simple closed curve C to a double integral over the plane region D bounded by C The theorem states:

C(Pdx+Qdy)=D(QxPy)dA

Calculations:

the area enclosed by C, we can use a specific form of Green's Theorem. Consider the vector field (P, Q) =(x/2. y/2). Applying Green's Theorem to this vector field gives us:

C(y2dx+x2dy)=D(x(x2)y(y2))dA

Calculating the partial derivatives, we get:

x(x2)=12

y(y2)=12

So, the integrand becomes:

D(12(12))dA=D(1)dA=DdA

The right-hand side is simply the area of D. Therefore, the left-hand side gives us the area enclosed by C

C(y2dx+x2dy)=Area(D)

Multiplying through by 2, we get:

C(ydx+xdy)=2Area(D)

Thus, the area A enclosed by C is:

A=12C(xdyydx)

Conclusion:

Therefore, the correct formula for the area of the region enclosed by a simple closed curve C is:

A=12C(xdyydx)

Among the given options, the correct answer is: 2

Green's Theorem Question 3:

Evaluate ∮c y3dx - x3dy where c is positively oriented circle of radius 2 centered at origin.

  1. 24Π
  2. 4Π
  3. (-24Π)
  4. (-4Π)

Answer (Detailed Solution Below)

Option 3 : (-24Π)

Green's Theorem Question 3 Detailed Solution

Calculations:

In this particular case, it's cleaner to apply Green's Theorem in polar coordinates. The equality of the line integral around the positively oriented circle of radius 2 centered at origin to a double integral over the region D enclosed by the curve is as follows:

∮C P dx + Q dy = ∫∫D (dQ/dx - dP/dy) dA

We first need to compute dQ/dx and dP/dy:

dQ/dx = d(-x3)/dx = -3x2 dP/dy = d(y3)/dy = 3y2

Then, dQ/dx - dP/dy = -3x2 - 3y2

In polar coordinates, x = rcos(θ) and y = rsin(θ), and the area element dA in polar coordinates is r dr dθ.

Replacing x and y with these and simplifying gives:

dQ/dx - dP/dy = -3r2(cos2(θ) + sin2(θ)) = -3r2.

So, by Green's Theorem, the line integral ∮C P dx + Q dy is equal to the double integral

∫ (from 0 to 2π) ∫ (from 0 to 2) -3r2 ×  r dr dθ.

Compute this double integral to obtain the final result. Let's do this:

= ∫ (from 0 to 2π) [(-3/4 ×  r4) from 0 to 2] dθ = ∫ (from 0 to 2π) (-3/4 × 16) dθ = -12 ×  ∫ (from 0 to 2π) dθ = -12 ×  [θ from 0 to 2π] = -12 ×  2π = -24π.

The value of ∮C y3 dx - x3 dy around the given circle in the positive direction is -24π.

Green's Theorem Question 4:

Pick the correct statement?

  1. Stoke's theorem is particular form of Green's theorem
  2. Green's theorem and Stoke's theorem are entirely different
  3. Green's theorem and Stoke's theorem are same
  4. Green's theorem is particular form of Stoke's theorem

Answer (Detailed Solution Below)

Option 4 : Green's theorem is particular form of Stoke's theorem

Green's Theorem Question 4 Detailed Solution

Solution - Green's Theorem - 

Let C be a positively orientedpiecewise smoothsimple closed curve in a plane,

and let D be the region bounded by C.

If L and M are functions of (xy) defined on an open region containing D and have continuous partial derivatives there, then

 

 

where the path of integration along C is anticlockwise. 

Stoke's Theorem - 

 

Let  be a smooth oriented surface in  with boundary . If a vector field  is defined and has continuous first order partial derivatives in a region containing , then

 

Here, Green's Theorem is a Particular form of Stoke's Theorem . 

Therefore, Correct Option is Option 4).

Green's Theorem Question 5:

Which theorem is valid in calculating integral for a multiple connected domain R?

  1. Green's theorem
  2. Stokes theorem
  3. Iterated integrals
  4. Gauss divergence theorem

Answer (Detailed Solution Below)

Option 1 : Green's theorem

Green's Theorem Question 5 Detailed Solution

Explanation:

Let C be the positively oriented, smooth, and simple closed curve in a plane, and D be the region bounded by the C. If P and Q are the functions of (x, y) defined on the open region, containing D and have continuous partial derivatives, then Green’s theorem is stated as

C(Pdx+Qdy) = D(QxPy)dxdy

Therefore, Green's theorem is valid in calculating integral for a multiple connected domain R.

Option (1) is true.

Green's Theorem Question 6:

The value of C[(cosxsinyxy)dx+sinxcosydy] where C is the circle x2 + y2 = 1, is:

  1. 0
  2. 1
  3. π/2
  4. π 

Answer (Detailed Solution Below)

Option 1 : 0

Green's Theorem Question 6 Detailed Solution

Concept:

Green's Theorem:-  If two functions M(x, y) and N(x, y and their partial derivatives are single valued and continuous over a region R  bounded by a closed curve C, then

(Mdx+Ndy)=R(NxMy)dxdy

Green Theorem is useful for evaluating a line integral around a closed curve C.

Calculation:

We have,

⇒ C[(cosxsinyxy)dx+sinxcosydy]

On comparing, we get

⇒ M = cos x sin y - x y

⇒ N = sin x cos y

On differentiating M partially with respect to 'y'

⇒ My=cosx cosyx

On differentiating N partially with respect to 'x'

⇒ Nx=cosx cosy

⇒ s[(cosx cosy)(cosxcosyx)]dxdy

⇒ sxdxdy 

⇒ x=11y=1x21x2   xdxdy

⇒ 11x[y]1x21x2dx

⇒ 11xdx[1x2+1x2]

⇒ 211x1x2dx

⇒ Let 1 - x2 = t

⇒ On differentiating, we get

⇒ - 2x dx = dt

⇒ x dx = dt2

When x = -1

⇒ 1 - 1 = 0 = t

When x = 1

⇒ 1 - 1 = 0 = t

⇒ 200tdt2

⇒ 2 × 0

⇒ 0

∴ The value of C[(cosxsinyxy)dx+sinxcosydy] where C is the circle x2 + y2 = 1 is 0.

Green's Theorem Question 7:

The value of ∫ xy dy – y2dx, where C is rectangle cut from the first quadrant by the lines x = 1 and y = 2 is

  1. 3/2
  2. 3
  3. 6
  4. 12

Answer (Detailed Solution Below)

Option 3 : 6

Green's Theorem Question 7 Detailed Solution

Concept:

By Green’s theorem

I=CMdx+Ndy=A(NxMy)dxdy(1)

Calculation:

Given:

M =  xy, N = – y2

NxMy=y(2y)=3y

By using equation (1),

I=xydyy2dx=x=01y=02(3y)dydx

I=x=0132[y2]02dx

 

I=016dx

Cxydyy2dx=6

Green's Theorem Question 8:

The value of the integral c[(2x2y2)dx+(x2+y2)dy] where C is the boundary of the area enclosed by the axis and the upper half of the circle x2 + y2 = a2 . Take a = 3.

Answer (Detailed Solution Below) 72

Green's Theorem Question 8 Detailed Solution

By using Green’s Theorem use have

c(Pdx+Qdy)=s[QxPy]dxdy

D172

c(2x2y2)dx+(x2+y2)dy

=aa0a2x2[x(x2+y2)y(2x2y2)]dxdy

=aa0a2x2[2x+2y]dxdy

=2aa0a2x2(x+y)dxdy

=2aadx0a2x2(xy+y22)dy

=2aadx[xa2x2+a2x22]

=2aa(xa2x2)dx+aa(a2x2)dx

0af(x)dx=={20af(x)dx,iffiseven0,iffisodd

=2[0+20a(a2x2)dx]

=4[a2xx33]0a

=4[a3a33]

l=4×2a33=8a33

With a = 3, we get the integrated value as 

l = 72

Green's Theorem Question 9:

Find the value of c[(xy+y2)dx+x2dy], where C is bounded by y = x and y = x2.

  1. 120
  2. 120
  3. 110
  4. 110

Answer (Detailed Solution Below)

Option 1 : 120

Green's Theorem Question 9 Detailed Solution

According to Green’s theorem,

c(Mdx+Ndy)=E(NxMy)dxdy

Here in the given question,

M = xy + y2

My=x+2y

N = x2

Nx=2x

C[(xy+y2)dx+x2dy]=E(2xx2y)dxdy=E(x2y)dxdy

correction diagram

E(x2y)dxdy=x=01y=x2x(x2y)dxdy=01[xyy2]x2xdx=01[x4x3]dx=[x55x44]01=120

Green's Theorem Question 10:

Green's theorem says that the total "microscopic circulation" in D is equal to the circulation around the boundary c = dD̅

  1. ∫ -F.ds
  2. ∫ F.ds 
  3. F.ds
  4. ∫c F.ds

Answer (Detailed Solution Below)

Option 4 : ∫c F.ds

Green's Theorem Question 10 Detailed Solution

Explanation:

Green's theorem states that the circulation form of the double integral over a plane region D is equal to the line integral around the boundary curve C of D (often noted as ∂D). To put it quantitatively:

∫∫_D (curl F) × da = ∮_C F ×  ds

Where:

The left side is a double integral over the region D of the curl of a vector field F dotted with da, the area element. This represents the total "microscopic circulation" in D.
The right side is a line integral around the closed curve C (the boundary of D, ∂D) of F dotted with ds, the line element. This represents the circulation around the boundary C.

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