Divergence Theorem MCQ Quiz in मराठी - Objective Question with Answer for Divergence Theorem - मोफत PDF डाउनलोड करा
Last updated on Mar 22, 2025
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Divergence Theorem Question 1:
Evaluate the surface integral \(\mathop{\int }_{S}\vec{F}\cdot ds\) when \(\vec{F}\) is given by \(\vec{F}=2xy~\hat{i}+8xz~\hat{j}+4yz~\hat{k}\) and S is the surface of tetrahedron whose vertices are (0, 0, 0), (1, 0, 0), (0, 1, 0) and (0, 0, 1).
Answer (Detailed Solution Below) 0 - 0.5
Divergence Theorem Question 1 Detailed Solution
Concept:
Gauss-divergence theorem:
\(\mathop{\int }_{S}\vec{F}\cdot ds=\iiint{\left( \vec{\nabla }\cdot \vec{F} \right)dV}\)
The tetrahedron is given as
Equation of xy is
\(\frac{{y - {y_x}}}{{{y_4} - {y_x}}} = \frac{{x - {X_x}}}{{{x_4} - {X_x}}} \Rightarrow y = 1 - x\)
Equation of plane XYZ
\(\frac{x}{1} + \frac{y}{1} + \frac{z}{1} = 1\)
x + y + z = 1
Calculation:
\(\vec \nabla \cdot \vec F = \left( {\frac{\partial }{{\partial x}}\hat i + \frac{\partial }{{\partial y}}\hat j + \frac{\partial }{{\partial z}}\hat k} \right) \cdot \left( {2xy\;\hat i + 8xz\;\hat j + 4yz\;\hat k} \right)\)
= 2y + 0 + 4y = 6y
∴ Integral is 6 \(\iiint{y~dx~dy~dz~}\)
\( = 6\mathop \smallint \limits_0^1 dx\mathop \smallint \limits_0^{1 - x} ydy\mathop \smallint \limits_0^{1 - x - y} dz\)
\(= 6\mathop \smallint \limits_0^1 dx\mathop \smallint \limits_0^{1 - x} y\left( {1 - x - y} \right)dy\)
\(= 6\mathop \smallint \limits_0^1 dx\mathop \smallint \limits_0^{1 - x} \left( {y - xy - {y^2}} \right)dy\)
\(= 6\mathop \smallint \limits_0^1 \frac{{{{\left( {1 - x} \right)}^3}}}{6}dx = - \left[ {\frac{{{{\left( {1 - x} \right)}^4}}}{4}} \right]_0^1\;\)
\(= \frac{1}{4} = 0.25\)Divergence Theorem Question 2:
Find a unit normal to the surface x2y + 2xz = 4 at (2, -2, 3)
Answer (Detailed Solution Below)
Divergence Theorem Question 2 Detailed Solution
ϕ = x2y + 2xz
Grad ϕ is a vector normal to the surface ϕ = constant
\(\nabla \phi = \left( {\frac{\partial }{{\partial {\rm{x}}}}{\rm{\hat i}} + \frac{\partial }{{\partial {\rm{y}}}}{\rm{\hat j}} + \frac{\partial }{{\partial {\rm{z}}}}{\rm{\hat k}}} \right)\left( {{{\rm{x}}^2}{\rm{y}} + 2{\rm{xz}}} \right)\)
∇ϕ = î (2xy + 2z) + ĵ (x2) + k̂ (2x)
\(\begin{array}{l} {\left. {\nabla \phi } \right|_{\left( {2, - 2,3} \right)}} = - 2{\rm{\hat i}} + 4{\rm{\hat j}} + 4{\rm{\hat k}}\\ \left| {\nabla \phi } \right| = \sqrt {4 + 16 + 16} = 6 \end{array}\)
Unit normal to the given surface at (2, -2, 3)
\(\frac{{\nabla \phi }}{{\left| {\nabla \phi } \right|}} = \frac{{ - 2{\rm{\hat i}} + 4{\rm{\hat j}} + 4{\rm{\hat k}}}}{6} = \frac{1}{3}\left( { - {\rm{\hat i}} + 2{\rm{\hat j}} + 2{\rm{\hat k}}} \right)\)Divergence Theorem Question 3:
The direction of vector A is radially outward from the origin, with \(\left| A \right| = k{r^n}\) where \({r^2} = {x^2} + {y^2} + {z^2}\) and k is a constant. The value of n for which \(\nabla \cdot {\rm{A}} = 0\) is
Answer (Detailed Solution Below) -2
Divergence Theorem Question 3 Detailed Solution
Divergence of A in spherical coordinates is given as
\(\nabla \cdot {\rm{A}} = \frac{1}{{{{\rm{r}}^2}}}.\frac{\partial }{{\partial {\rm{\;r}}}}\left( {{{\rm{r}}^2}{{\rm{A}}_{\rm{r}}}} \right) = \frac{1}{{{{\rm{r}}^2}}}\frac{\partial }{{\partial {\rm{\;r}}}}\left( {{\rm{k}}{{\rm{r}}^{{\rm{n}} + 2}}} \right)\)
\( = \frac{{\rm{k}}}{{{{\rm{r}}^2}}}\left( {{\rm{n}} + 2} \right){{\rm{r}}^{{\rm{n}} + 1}}\)
Given that \(\nabla \cdot {\rm{A}} = 0\)
\( \Rightarrow {\rm{k}}\left( {{\rm{n}} + 2} \right){{\rm{r}}^{{\rm{n}} - 1}} = 0\)
\( \Rightarrow {\rm{n}} + 2 = 0{\rm{\;}} \Rightarrow {\rm{n}} = - 2\)
Divergence Theorem Question 4:
Find the value of \(\mathop \smallint \limits_s \vec F.\hat nds\), if \(F = \left( {x + 3y} \right)\hat i + \left( {y + 2z} \right)\hat j + \left( {x - 2z} \right)\hat k\)
Answer (Detailed Solution Below) 0
Divergence Theorem Question 4 Detailed Solution
Using divergence theorem
\({\rm{}}\mathop \oint \limits_s \vec {F.}\hat nds = \mathop \smallint \limits_v \nabla .F\ dv\)
\(\nabla .F = \left( {\hat i\frac{\partial }{{\partial x}} + \hat j\frac{\partial }{{\partial y}} + \hat k\frac{\partial }{{\partial z}}} \right).\left[ {\left( {x + 3y} \right)\hat i + \left( {y + 2z} \right)\hat j + \left( {x - 2z} \right)\hat k} \right]\)
\(\rm = 1 + 1 – 2 = 0\)
So, \(\mathop \oint \nolimits \widehat {F.}\hat nds = 0\)
Divergence Theorem Question 5:
\(V = \left\{ {\left( {x,y,z} \right)\epsilon{R^3},:1 \le {x^2} + {y^2} + {z^2} \le 4} \right\}\) find the value of \(\mathop \smallint \limits_s \vec F.\hat nds\), if \(F = \left( {x + 3y} \right)\hat l + \left( {y + 2z} \right)\hat j + \left( {x - 2z} \right)\hat k\)
Answer (Detailed Solution Below) 0
Divergence Theorem Question 5 Detailed Solution
Using divergence theorem
\(\begin{array}{l} {\rm{}}\mathop \oint \limits_s \widehat {F.}\hat nds = \mathop \smallint \limits_v \nabla .F\ dv\\ \nabla .F = \left( {\hat l\frac{\partial }{{\partial x}} + \hat j\frac{\partial }{{\partial y}} + \hat k\frac{\partial }{{\partial z}}} \right).\left[ {\left( {x + 3y} \right)\hat l + \left( {x + 2z} \right)\hat j + \left( {x + 2z} \right)\hat k} \right] \end{array}\)
= 1 + 1 – 2 = 0
So, \(\mathop \oint \nolimits \widehat {F.}\hat nds = 0\)