Divergence MCQ Quiz in मराठी - Objective Question with Answer for Divergence - मोफत PDF डाउनलोड करा

Divergence of a function \(F = {F_1}\hat i + {F_2}\hat j + {F_3}\hat k\)

\( = \nabla \cdot F = \frac{{\partial {F_1}}}{{\partial x}} + \frac{{\partial {F_2}}}{{\partial y}} + \frac{{\partial {F_3}}}{{\partial z}}\)

Explanation:

Divergence of a vector function \(F = {F_1}\hat i + {F_2}\hat j + {F_3}\hat k\)

\( = \nabla \cdot F = \frac{{\partial {F_1}}}{{\partial x}} + \frac{{\partial {F_2}}}{{\partial y}} + \frac{{\partial {F_3}}}{{\partial z}}\)

Calculation:

Vector = -yi +xj

Concept:

The Divergence theorem states that:

\(\mathop{{\int\!\!\!\ \!\!\int}\mkern-21mu \ \bigcirc} {D.ds = \iiint_V {\left( {∇ .D} \right)\;dV}}\)

where ∇.D is the divergence of the vector field D.

In Rectangular coordinates, the divergence is defined as:

\(\nabla \cdot \vec D = \left( {\frac{{\partial {D_x}}}{{\partial x}} + \frac{{\partial {D_y}}}{{\partial y}} + \frac{{\partial {D_z}}}{{\partial z}}} \right)\)

Analysis:

\(\nabla{\rm{\;}}\left( {{\rm{xi}} + {\rm{yj}} + {\rm{zk}}} \right) \)

\(= \frac{\partial }{{\partial {\rm{x}}}}\left( {\rm{x}} \right) + \frac{\partial }{{\partial {\rm{y}}}}\left( {\rm{y}} \right) + \frac{\partial }{{\partial {\rm{z}}}}\left( {\rm{z}} \right) = 3\)

• Divergence operates on a vector field but results in a scalar.
• Curl operates on a vector field and results in a vector field.
• Gradient operates on a scalar but results in a vector field.
• Divergence of curl, Curl of the gradient is always zero.
• Thus, the gradient of curl gives the result of curl (which is a vector field) to the gradient to operate upon, which is a mathematically invalid expression.

We have

\(\vec \nabla .\vec V = {1 \over {{r^2}}}{\partial \over {\partial r}}\left( {{r^2}.{V_r}} \right) + {1 \over {rsin\theta }}{\partial \over {\partial \theta }}\left( {sin\theta {V_\theta }} \right) + {1 \over {rsin\theta }}\left( {{{\partial {V_\phi }} \over {{\partial _\phi }}}} \right)\)

\(\vec \nabla .\vec V = {1 \over {{r^2}}} \hat{r} + 0. \hat {\theta} + 0.\hat {\phi} \)

Thus,

\(\vec \nabla .\vec V = {1 \over {{r^2}}}{\partial \over {\partial r}}\left( {{r^2}.{V_r}} \right) = {1 \over {{r^2}}}{\partial \over {\partial r}}\left( 1 \right) = 0\)

Explanation:

If a vector point function F(x, y, z) = F₁i  + F₂j + F₃k is defined and differentiable at each point in some region of space then the divergence of F is denoted by div F or ∇.F and define as,

div F = ∇.F = \(\frac{{\delta {F_1}}}{{\delta x}} + \frac{{\delta {F_2}}}{{\delta y}} + \frac{{\delta {F_3}}}{{\delta z}}\) 

Solenoidal Vector: A vector function F is said to be a solenoidal vector if ∇.F = 0.

Additional Information

Irrotational Vector: ​A vector point function F is said to be a rotational vector if curl F = 0

curl F = ∇ × F = \(\left[ {\begin{array}{*{20}{c}} i&j&k\\ {\frac{\delta }{{\delta x}}}&{\frac{\delta }{{\delta y}}}&{\frac{\delta }{{\delta z}}}\\ {{F_1}}&{{F_2}}&{{F_3}} \end{array}} \right]\)

Scalar potential vector: If for a given irrotational vector F there exist a scalar point function ϕ (x, y,z) such that F = ∇ ϕ then ϕ (x, y, z) is called a scalar potential function of F.

Concept:

Divergence of a vector:

Divergence of any vector \(\vec{A}=F_1̂ i + F_2̂ j+ F_3̂ k\) is given by:

\(\nabla.\vec {A}=\left(̂{i}\frac{\partial}{\partial x}+̂{j}\frac{\partial}{\partial y}+̂{k}\frac{\partial}{\partial z}\right).(F_1̂ i + F_2̂ j+ F_3̂ k)\)

\(\nabla.\vec {A}=\left(\frac{\partial F_1}{\partial x}+\frac{\partial F_2}{\partial y}+\frac{\partial F_3}{\partial z}\right)\)

Calculation:

Given:

A vector to the solenoidal, divergence of that vector should be zero.

F = (ax²y + yz)î + (xy² - xz²)ĵ + (2xyz - 2x²y²)k̂ 

∇.F = 2axy + 2xy + 2xy = 0

Concept:

The divergence of any vector field  \(\vec A\) is defined as:

\(Div= \vec \nabla .\vec A\)

The nabla operator is defined as:

\(\vec \nabla ={\hat i\frac{\partial }{{\partial x}} + \hat j\frac{\partial }{{\partial y}}}+\hat k\frac{\partial }{\partial z}\)

Calculation:

\(f = {x^3}y\vec i - \left( {{z^2} - 2y} \right)\vec j + 5{y^2}z\vec k\)

\(Div\left( f \right) = \frac{\partial }{{\partial x}}\left( {{x^3}y} \right) + \frac{\partial }{{\partial y}}\left( { - \left( {{z^2} - 2y} \right)} \right) + \frac{\partial }{{\partial z}}\left( {5{y^2}z} \right)\)

= 3x²y + (+2) + (5y²)

= 3x²y + 5y² + 2

At (2, 3, 4)

Div (f) = 3(2)²(3) + 5(3)² + 2

∴ Div (f) = 36 + 45 + 2 = 83

Concept:

Divergence of the vector function:

The net outward flux from a volume element around a point is a measure of the divergence of the vector field at that point.

Divergence of a function \(\vec F = {F_1}\hat i + {F_2}\hat j + {F_3}\hat k\)

\( \nabla \cdot \vec F = \frac{{\partial {F_1}}}{{\partial x}} + \frac{{\partial {F_2}}}{{\partial y}} + \frac{{\partial {F_3}}}{{\partial z}}\)

Calculation:

Given that,

\(\vec{f}\ =\ 6x^2\hat{i}\ +\ 3xy^2\hat{j}\ +\ xyz^3\hat{k}\)

Therefore, the divergence of a given vector is

\( \nabla \cdot \vec f = \frac{{\partial {f_1}}}{{\partial x}} + \frac{{\partial {f_2}}}{{\partial y}} + \frac{{\partial {f_3}}}{{\partial z}}\)        -----(1)

Here,

f₁ = 6x², f₂ = 3xy², f₃ = xyz³      

Therefore, from equation (1)

\( \nabla \cdot \vec f = \frac{{\partial {}}}{{\partial x}}6x^2 + \frac{{\partial {}}}{{\partial y}}3xy^2 + \frac{{\partial {}}}{{\partial z}}xyz^3\)

⇒ \( \nabla \cdot \vec f \) = 12x + 6xy + 3xyz²

Therefore, divergence at point (2, 3, 4)

\( \nabla \cdot \vec f \) = 12(2) + 6(2)(3) + 3(2)(3)(4)²

⇒ \( \nabla \cdot \vec f \)  = 24 + 36 + 288 = 348

Hence, option 3 is correct.

Additional Information

Gauss divergence theorem: 

It states that the surface integral of the normal component of a vector function \(\vec F\) taken over a closed surface ‘S’ is equal to the volume integral of the divergence of that vector function \(\vec F\) taken over a volume enclosed by the closed surface ‘S’.

\(\underset{S}{\mathop \iint }\,\vec{F}.\hat{n}ds=\iiint\limits_{V}{\nabla .\vec{F}dv}\)

Explanation:

\(\nabla \cdot \left( {\phi \vec A} \right) = \left( {\nabla \phi } \right) \cdot \vec A + \phi \left( {\nabla \cdot \vec A} \right)\)

\(\nabla \cdot \left( {\frac{1}{{{r^2}}} \cdot \vec r} \right) = \left( {\nabla \frac{1}{{{r^2}}}} \right) \cdot \vec r + \frac{1}{{{r^2}}}\left( {\nabla \cdot \vec r} \right)\)

\(= \left[ {\frac{{ - 2\vec r}}{{{r^4}}}} \right] \cdot \vec r + \frac{3}{{{r^2}}}\;\;\;\;\;\;\;\;\;\;\;\left[ {\because \nabla \;\left( {f\left( r \right)} \right) = \frac{{f'\left( r \right)}}{r}\vec r} \right]\)

\( = \frac{{ - 2}}{{{r^2}}} + \frac{3}{{{r^2}}}\;\;\;\;\;\;\;\;\;\;\left\{ {\vec r \cdot \vec r = {r^2}} \right\}\)

\(= \frac{1}{{{r^2}}}\)

\({\nabla ^2}\left[ {\nabla \cdot \frac{{\vec r}}{{{r^2}}}} \right] = {\nabla ^2}\left[ {\frac{1}{{{r^2}}}} \right]\)

\( = \frac{6}{{{r^4}}} + \frac{2}{r}\left( {\frac{{ - 2}}{{{r^3}}}} \right)\)

\(\left[ {{\nabla ^2}\left\{ {f\left( r \right)} \right\} = r''\left( r \right) + \frac{2}{r}f'\left( r \right)} \right]\)

\(= \frac{6}{{{r^4}}} - \frac{4}{{{r^4}}}\)

\(= \frac{2}{{{r^4}}} = 2{r^{ - 4}}\)

Calculation:

For an Incompressible flow, div \(\vec V = 0\)

\(div\;\left( {\vec V} \right) = \vec \nabla \cdot \vec V = \frac{\partial }{{\partial x}}\left( {2x + 3y + 4z} \right) + \frac{{\partial \left( {5x + cy + 6z} \right)}}{{\partial y}} + \frac{\partial }{{\partial z}}\left( {8x + 9y} \right){\rm{\;}} = {\rm{\;}}2{\rm{\;}} + {\rm{\;C}}\)

For incompressible flow, \(\vec \nabla \cdot \vec V = 0\)

⇒ 2 + C = 0 ⇒ C = -2