Curl MCQ Quiz - Objective Question with Answer for Curl - Download Free PDF

Explanation:

v . curl v = v . ∇ × v = [v ∇ v] = 0

Option (4) is true. 

Explanation:

v . curl v = v . ∇ × v = [v ∇ v] = 0

Option (4) is true. 

Explanation:

\(\rm\vec a\) is differentiable vector point function and u is a differentiable scalar point function.

Then \(\rm \nabla× (u\vec a)\)

  = curl × \((u\vec a)\)

  = \(\rm (\nabla u)\times \vec a +u(\nabla \times \vec a)\)

Option (1) is true.

Explanation:

⇒ Since \(\mathbf{F} \)  is irrotational, we have \(\nabla \times \mathbf{F} = \mathbf{0} \) 

The curl of \(\mathbf{F} \)  is given by : 

⇒\(\nabla \times \mathbf{F} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ P & Q & R \end{vmatrix}\)

⇒ where P = x + 2y + az,  Q = bx - 3y - z,   R = 4x + cy + 2z. 

⇒\(\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ x + 2y + az & bx - 3y - z & 4x + cy + 2z \end{vmatrix}\)

⇒ Expanding the determinant

⇒\(\nabla \times \mathbf{F} = \hat{i} \begin{vmatrix} \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ bx - 3y - z & 4x + cy + 2z \end{vmatrix} - \hat{j} \begin{vmatrix} \frac{\partial}{\partial x} & \frac{\partial}{\partial z} \\ x + 2y + az & 4x + cy + 2z \end{vmatrix} \)\(+ \hat{k} \begin{vmatrix} \frac{\partial}{\partial x} & \frac{\partial}{\partial y} \\ x + 2y + az & bx - 3y - z \end{vmatrix}\)

⇒ \( \nabla \times \mathbf{F} = (c + 1) \hat{i} - (4 - a) \hat{j} + (b - 2) \hat{k} \)

Since \(\mathbf{F} \)  is irrotational

c + 1 = 0,  4 - a = 0,  b - 2 = 0 

⇒\(c = -1, \quad a = 4, \quad b = 2\)

⇒\(a^2 + b^2 + c^2 = 4^2 + 2^2 + (-1)^2 = 16 + 4 + 1 = 21\)

Hence option 4 is correct

If  \(\vec{r}=x \vec{\imath}+y \vec{\jmath}+z \vec{k}\) & \(\vec{a}=a_1 \vec{\imath}+a_2 \vec{\jmath}+a_3 \vec{k}\) Then \(\vec{r} \times \vec{a}=\begin{bmatrix} ​​​​\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ {x} & {y} & {z} \\ a_1 & a_2 & a_3 \end{bmatrix} \)

Explanation:

\(\vec{r}=x \vec{\imath}+y \vec{\jmath}+z \vec{k}\) & \(\vec{a}=a_1 \vec{\imath}+a_2 \vec{\jmath}+a_3 \vec{k}\)

Then \(\vec{r} \times \vec{a}=\begin{bmatrix} ​​​​\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ {x} & {y} & {z} \\ a_1 & a_2 & a_3 \end{bmatrix} \)

= \(\rm \hat{\mathbf{i}}(a_3y-a_2z)-\hat{\mathbf{j}}(a_3x-a_1z)+\hat{\mathbf{k}}(a_2x-a_1y)\)

Now, \(\nabla \times(\vec{r} \times \vec{a})=​​\)\(\begin{bmatrix} ​​​​\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ \frac{\partial}{\partial{x}} & \frac{\partial}{\partial{y}} & \frac{\partial}{\partial{z}} \\ (a_3y-a_2z) & (a_1z-a_3x) & (a _2x-a_1y) \end{bmatrix} \)

⇒ \(\nabla \times(\vec{r} \times \vec{a})=​​(-a_1-a_1)\hat{\mathbf{i}}-(a_2+a_2)\hat{\mathbf{j}}+(-a_3-a_3)\hat{\mathbf{k}}\)

⇒ \(\nabla \times(\vec{r} \times \vec{a})=​​-2a_1\hat{\mathbf{i}}-2a_2\hat{\mathbf{j}}+-2a_3\hat{\mathbf{k}}\)

⇒ \(\nabla \times(\vec{r} \times \vec{a})=​​-2a\)

As given \(\nabla \times(\vec{r} \times \vec{a})=m \vec{a}\)

⇒ m= -2

Concept:

The curl of a vector is given by the expansion of the following matrix

If \(\vec{f}={{f}_{1}}\hat{i}+{{f}_{2}}\hat{j}+{{f}_{3}}\hat{k}\)

Then

\(\nabla \times ~\vec{f}=\left| \begin{matrix} {\hat{i}} & {\hat{j}} & \hat{k} \\ \frac{\partial }{\partial x} & \frac{\partial }{\partial y} & \frac{\partial }{\partial z} \\ {{f}_{1}} & {{f}_{2}} & {{f}_{3}} \\ \end{matrix} \right|\)

Calculation:

Given vector is

\(~\vec{u}=yz~\hat{i}+3zx~\hat{j}+z~\hat{k}\)

Than \(\nabla \times ~\vec{u}=\left| \begin{matrix} {\hat{i}} & {\hat{j}} & {\hat{k}} \\ \frac{\partial }{\partial x} & \frac{\partial }{\partial y} & \frac{\partial }{\partial z} \\ yz & 3zx & z \\ \end{matrix} \right|\)

\(\nabla \times ~\vec{u}=\hat{i}\left( \frac{\partial (z)}{\partial y}-\frac{\partial (3zx)}{\partial z} \right)-\hat{j}\left( \frac{\partial (z)}{\partial x}-\frac{\partial (yz)}{\partial z} \right)+~\hat{k}\left( \frac{\partial (3zx)}{\partial x}-\frac{\partial( yz)}{\partial y} \right)\)

Concept:

For a vector F = F₁i + F₂j + F₃k

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

\(Curl = \nabla \times F = \left| {\begin{array}{*{20}{c}} i&j&k\\ {\frac{\partial }{{\partial x}}}&{\frac{\partial }{{\partial y}}}&{\frac{\partial }{{\partial z}}}\\ {{F_1}}&{{F_2}}&{{F_3}} \end{array}} \right|\)

For irrotational (or) conservative field \(\nabla \times \vec F = 0\) (or) Null Vector.

Calculation:

Given that,

\(\vec F = {\hat a_x}\;\left( {5y - {k_1}z} \right) + {\hat a_y}\left( {3z + {k_2}x} \right) + {\hat a_z}\left( {{k_3}y - 4x} \right)\) is a conservative field.

\(\left| {\begin{array}{*{20}{c}} {{{\hat a}_x}}&{{{\hat a}_y}}&{{{\hat a}_z}}\\ {\frac{\partial }{{\partial x}}}&{\frac{\partial }{{\partial y}}}&{\frac{\partial }{{\partial z}}}\\ {\left( {5y - {k_1}z} \right)}&{\left( {3z + {k_2}x} \right)}&{\left( {{k_3}y - 4x} \right){\rm{\;}}} \end{array}} \right| = 0\)

\( \Rightarrow {\hat a_x}\;\left( {{{\rm{k}}_3} - 3} \right) - {\hat a_y}\left( {\; - 4 + {{\rm{k}}_1}} \right) + {\hat a_z}\left( {{{\rm{k}}_2} - 5} \right)\)

k₃ – 3 = 0 ⇒ k₃ = 3

-4 + k₁ = 0 ⇒ k₁ = 4

k₂ – 5 = 0 ⇒ k₂ = 5

Given that, \(\vec A = \vec \nabla u\)

\(\left| {\vec \Delta \times \vec A} \right| = curl\left( {\vec A} \right)\)

\(= curl\left( {\vec \nabla {\rm{u}}} \right)\)

We know that, for any scalar function ‘f’

\(curl\;\left( {\vec \nabla {\rm{f}}} \right) = 0\)

Concept:

Consider the vector field

\(\vec V = 2{x^2}yi + 5{z^2}j - 4yzk\)

\(curl\;\vec V = \nabla \times \vec V = \left| {\begin{array}{*{20}{c}} i&j&k\\ {\frac{\partial }{{\partial x}}}&{\frac{\partial }{{\partial y}}}&{\frac{\partial }{{\partial z}}}\\ {2{x^2}y}&{5{z^2}}&{ - 4yz} \end{array}} \right|\)

\(\Rightarrow Curl\;\vec V = i\left[ {\frac{\partial }{{\partial y}}\left( { - 4yz} \right) - \frac{\partial }{{\partial z}}\left( {5{z^2}} \right)} \right] - j\left[ {\frac{\partial }{{\partial x}}\left( { - 4yz} \right) - \frac{\partial }{{\partial z}}\left( {2{x^2}y} \right)} \right] + k\left[ {\frac{\partial }{{\partial x}}\left( {5{z^2}} \right) - \frac{\partial }{{\partial y}}\left( {2{x^2}y} \right)} \right]\)

\(\Rightarrow Curl\;\vec V = i\left[ { - 4z - 10z} \right] - j\left[ 0 \right] + k\left[ { - 2{x^2}} \right]\)

Div (grad ϕ) = \(\vec{\nabla }\cdot \left( \vec{\nabla }\phi \right)=\left( \vec{\nabla }\cdot \vec{\nabla } \right)\cdot \phi ~={{\nabla }^{2}}\phi\)

Let ϕ be a function of (x, y, z)

Then grad \(\left( \phi \right)=i\frac{\partial \phi }{\partial x}+j\frac{\partial \phi }{\partial y}+k\frac{\partial \phi }{\partial z}\)

Divergence (grad ϕ) \(=\frac{{{\partial }^{2}}\phi }{\partial {{x}^{2}}}+\frac{{{\partial }^{2}}\phi }{\partial {{y}^{2}}}+\frac{{{\partial }^{2}}\phi }{\partial {{z}^{2}}}\)

Concept:

Curl of a Vector:

Let \(\rm \vec{v} = v_1 \rm \vec{i} + v_2 \rm \vec{j}+ v_3 \rm \vec{k}\)

Then the curl of the vector function v is given by:

\(\rm curl \rm \;\vec v =∇ \times {\rm{\;\vec v\;}}=\begin{vmatrix} \rm \vec i & \rm \vec j & \rm \vec k \\ \frac{\partial }{\partial x} & \frac{\partial }{\partial y} & \frac{\partial }{\partial z} \\ \rm v_1 & \rm v_2 & \rm v_3 \end{vmatrix}\)

Where ∇ = \(\rm \frac{{\partial }}{{\partial x}} \hat{i}+ \frac{{\partial }}{{\partial y }} \hat{j} +\frac{{\partial }}{{\partial z }} \hat{k}\)

Calculation:

Given: \(\rm \vec v\) = yz î + 3xz ĵ + z k̂

\(\rm curl \rm \;\vec v =∇ \times {\rm{\;\vec v\;}}=\begin{vmatrix} \rm \vec i & \rm \vec j & \rm \vec k \\ \frac{\partial }{\partial x} & \frac{\partial }{\partial y} & \frac{\partial }{\partial z} \\ \rm yz & \rm 3xz & \rm z \end{vmatrix}\)

\(= \rm \vec i (\frac{\partial z}{\partial y} - \frac{\partial (3xz)}{\partial z})-\rm \vec j (\frac{\partial z}{\partial x} - \frac{\partial (yz)}{\partial z})+\rm \vec k (\frac{\partial (3xz)}{\partial x} - \frac{\partial (yz)}{\partial y})\)

\(= \rm \vec i (0-3x)- \rm \vec j (0-y)+\rm \vec k (3z-z)\\= -3x\rm \vec i +y \rm \vec j +2z\rm \vec k \)

Concept:

Let \(\vec A = {A_x}\;{a_x} + {A_y}{a_y} + {A_z}\;{a_z}\) 

The curl of A is evaluated as:

\(Curl\;\vec A = \left| {\begin{array}{*{20}{c}} {{a_x}}&{{a_y}}&{{a_z}}\\ {\frac{\partial }{{{\partial x}}}}&{\frac{\partial }{{{\partial y}}}}&{\frac{\partial }{{{\partial z}}}}\\ {{A_x}}&{{A_y}}&{{A_z}} \end{array}} \right|\)

Analysis:

Given

\(\vec A\) = (cos x) (sin y) a_x + (sin x) (cos y) â_y + 0⋅a_z

A_x = cos x sin y

A_y = sin x cos y

A_z = 0

\(\nabla \times \vec A = \left| {\begin{array}{*{20}{c}} {{a_x}}&{{a_y}}&{{a_z}}\\ {\frac{\partial }{{{\partial x}}}}&{\frac{\partial }{{{\partial y}}}}&{\frac{\partial }{{{\partial z}}}}\\ {\cos x\sin y}&{\sin x\cos y}&0 \end{array}} \right|\)

\(\nabla \times \vec A = {a_x}\left( {0 - 0} \right) - {a_y}\left( {0 - 0} \right) + {a_z}\left( {\cos x\cos y - \cos x\cos y} \right)\)

\(\nabla \times \vec A = 0\;{a_x} - 0\;{a_y} - 0\;{a_z}\)

\(\left| {\nabla \times \vec A} \right| = 0\)

Concept:

Let ϕ be a function of (x, y, z)

Then grad \(\left( ϕ \right)=\hat{i}\frac{\partial ϕ }{\partial x}+\hat{j}\frac{\partial ϕ }{\partial y}+\hat{k}\frac{\partial ϕ }{\partial z}\)

curl (grad (ϕ)) \(=\left| \begin{matrix} {\hat{i}} & {\hat{j}} & {\hat{k}} \\ \frac{\partial }{\partial x} & \frac{\partial }{\partial y} & \frac{\partial }{\partial z} \\ \frac{\partial ϕ }{\partial x} & \frac{\partial ϕ }{\partial y} & \frac{\partial ϕ }{\partial z} \\ \end{matrix} \right|\)

curl (grad (ϕ)) \(=0\hat{i}+0\hat{j}+0\hat{k}\)

Concept:

Gradient: Let a function F(x, y, z) then the gradient of F or ∇F is denoted as: 

∇F = \(\frac{∂F}{∂x}\hat i \ + \ \frac{∂F}{∂y}\hat j \ + \ \frac{∂F}{∂z} \hat k\), where i, j, k are unit vectors.

Curl: In 3 dimensional cartesian coordinate the curl of a vector field \(\vec F = \vec F_x \hat i \ + \ \vec F _y \hat j \ + \ \vec F_ z \hat k\) is defined as:-

curl \(\vec F\) or \(\vec \bigtriangledown × \vec F\) = \(\left( \frac{∂}{∂x}\hat i \ + \ \frac{∂}{∂y} \hat j \ + \ \frac{∂}{∂z} \hat k \right) × \left( \vec F_x \hat i \ + \ F_y \hat j \ + \ F_z \hat k \right)\)

\(\vec \bigtriangledown × \vec F\) = \( \left|\begin{matrix} \hat i & \hat j & \hat k \\ \frac{∂}{∂x} & \frac{∂}{∂y} & \frac{∂}{∂z} \\ \vec F_x & \vec F_y & \vec F_z \end{matrix} \right|\)

\(\vec \bigtriangledown × \vec F\) = \(\hat i \left( \frac{∂}{∂y} F_z - \frac{∂}{∂z}F_y \right) - \hat j \left( \frac{∂}{∂x} F_z - \frac{∂}{∂z}F_x \right) + \hat k \left( \frac{∂}{∂x}F_y - \frac{∂}{∂y} F_x \right)\)

Divergence: In 3 dimensional cartesian coordinate the divergence of a continuously differentiable vector field F = F_xi + F_yj + F_zk is defined as the scalar valued function.

It is denoted by divF = ∇.F

∇.F = \(\left( \frac{∂}{∂x} i + \frac{∂}{∂y}j + \frac{∂}{∂z} k \right) . (F_x i + F_y j + F_z k )\)

∇.F = \(\frac{∂F_x}{∂x} + \frac{∂F_y}{∂y} + \frac{∂F_z}{∂z}\)

Formula Used:

\(\vec a \times (\vec b \times \vec c ) = (\vec a .\vec c) \vec b - (\vec a.\vec b) \vec c\)

Calculation:

We have,

⇒ \(\vec \bigtriangledown × \vec F\) = \(\hat i × \frac{∂F}{∂x} + \hat j × \frac{∂F}{∂y} +\hat k × \frac{∂F}{∂z}\)

⇒ \(∇ × (∇ × \vec F)\) = \(\sum \hat i × \frac{∂}{∂x} \left(\hat i × \frac{∂F}{∂x} + \hat j × \frac{∂F}{∂y} +\hat k × \frac{∂F}{∂z} \right)\)

⇒ \(∇ × (∇ × \vec F)\) = \(\sum \hat i × \left[ i × \frac{∂^2\vec F}{∂x^2} + \hat j × \frac{∂^\vec 2F}{∂x∂y} +\hat k × \frac{∂^2\vec F}{∂x∂dz} \right]\)

⇒ \(∇ × (∇ × \vec F)\) = \(\sum \left[\hat i × \left( \hat i × \frac{∂^2\vec F}{∂x2} \right)+ \hat i × \left(\hat j × \frac{∂^\vec 2F}{∂x∂y} \right)+\hat i × \left(\hat k × \frac{∂^2\vec F}{∂x∂dz}\right) \right]\)

⇒ \(\sum \left[ \left(\hat i . \frac{∂^2 \vec F}{∂x^2} \right) \hat i - (\hat i .\hat i) \frac{∂^2 \vec F}{∂x^2} + \left( \hat i. \frac{∂^2 \vec F}{∂x∂dy}\right) \hat j - (\hat i .\hat j) \frac{∂^2\vec F}{∂x∂y} + \left( \hat i . \frac{∂^2\vec F}{∂x∂z} \right) \hat k - (\hat i.\hat k) \frac{∂^2\vec F}{∂x∂z} \right]\)

⇒ \(\sum \left[ \left(\hat i. \frac{∂^2\vec F}{∂x^2}\right) \hat i - \frac{∂^2 \vec F}{∂x^2} + \left(\hat i. \frac{∂^2\vec F}{∂x∂y} \right) \hat j - 0 + \left( \hat i. \frac{∂^2 \vec F}{∂x∂z}\right) \hat k - 0 \right]\)

⇒ \(\sum \left[ \hat i \frac{∂}{∂x} \left(\hat i. \frac{∂\vec F}{∂x}\right) + \hat j \frac{∂}{∂y}\left( \hat i . \frac{∂\vec F}{∂x}\right) + \hat k \frac{∂}{∂z}\left(\hat i . \frac{∂\vec f}{∂x}\right) \right] - \sum \frac{∂^2\vec F}{∂x^2}\)

⇒ \(\sum \left( \hat i \frac{∂}{∂x} + \hat j \frac{∂}{∂y} + \hat k \frac{∂}{∂z} \right) \left(\hat i . \frac{∂\vec F}{∂x}\right) - \left[ \frac{∂^2\vec F}{∂x^2} + \frac{∂^2\vec F}{∂y^2} + \frac{∂^2\vec F}{∂x^2}\right]\)

⇒ \(\sum ∇ \left( \hat i . \frac{∂\vec F}{∂x}\right) - ∇^2\vec F\)

⇒ \(∇ \left( \sum \hat i . \frac{∂\vec F}{∂x}\right) - ∇^2 \vec F\)

⇒ \(∇(∇.F) - ∇^2F\)

⇒ \(∇ \times (∇ \times \vec F) = ∇(∇.F) - ∇^2F\)

⇒ \(curl curl \vec F = grad \ div\ \vec F - ∇^2 \vec F\)

⇒ \(grad \ div \ \vec F = curl \ curl \vec F + ∇^2\vec F\)

∴ The true statement is \(grad \ div \ \vec F = curl \ curl \vec F + ∇^2\vec F\)

Concept:

If a vector function is given by

\(\vec \phi = u\widehat {i} + v\hat j + \omega \hat k\)

Then \(Curl~\vec \phi = \left| {\begin{array}{*{20}{c}} {\hat i}&{\hat j}&{\hat k}\\ {\frac{\partial }{{\partial x}}}&{\frac{\partial }{{\partial y}}}&{\frac{\partial }{{\partial z}}}\\ u&v&\omega \end{array}} \right|\)

Calculation:

Given:

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

\(Curl~\vec V = \left| {\begin{array}{*{20}{c}} {\hat i}&{\hat j}&{\hat k}\\ {\frac{\partial }{{\partial x}}}&{\frac{\partial }{{\partial y}}}&{\frac{\partial }{{\partial z}}}\\ {xyz}&{3{x^2}y}&{x{z^2} - {y^2}z} \end{array}} \right|\)

⇒ \(\left( { - 2yz} \right)\hat i - \left( {{z^2} - xy} \right)\hat j + \left( {6xy - xz} \right)\hat k\)

\({\left( {Curl~\vec V} \right)_{\left( {2, - 1,1} \right)}} = 2\hat i - 3\hat j - 14\hat k\)