Question 1 |
The vector function expressed by
F = a_x(5y - k_1z) + a_y(3z + k_2x) + a_z(k_3y - 4x)
represents a conservative field, where a_x, a_y, a_z are unit vectors along x, y and z directions, respectively. The values of constants k_1, k_2, k_3 are given by:
F = a_x(5y - k_1z) + a_y(3z + k_2x) + a_z(k_3y - 4x)
represents a conservative field, where a_x, a_y, a_z are unit vectors along x, y and z directions, respectively. The values of constants k_1, k_2, k_3 are given by:
k_1=3, k_2=3, k_3=7 | |
k_1=3, k_2=8, k_3=5 | |
k_1=4, k_2=5, k_3=3 | |
k_1=0, k_2=0, k_3=0 |
Question 1 Explanation:
\bar{F}=(5y-k_{1}Z)\hat{i}+(3z+k_{2}x)\hat{j}+(k_{3}y-4x)\hat{k}
is conservative field
\bar{F} is irrotational,
\begin{aligned} \triangledown \times \bar{F}&=0\\ \begin{vmatrix} \hat{i} &\hat{j} &\hat{k} \\ \frac{\partial }{\partial x}&\frac{\partial }{\partial y} &\frac{\partial }{\partial z} \\ 5y-k_{1}z &3z-k_{2}x &k_{3}y -4x \end{vmatrix}&=0\\ \hat{i}(k_{3}-3)-\hat{j}(-4+k_{1})+\hat{k}(k_{2}-5)&=0\\ k_{3}-3&=0\\ 4-k_{1}&=0\\ k_{2}-5&=0\\ k_{1}&=4\\ k_{2}&=5\\ k_{3}&=3 \end{aligned}
\bar{F} is irrotational,
\begin{aligned} \triangledown \times \bar{F}&=0\\ \begin{vmatrix} \hat{i} &\hat{j} &\hat{k} \\ \frac{\partial }{\partial x}&\frac{\partial }{\partial y} &\frac{\partial }{\partial z} \\ 5y-k_{1}z &3z-k_{2}x &k_{3}y -4x \end{vmatrix}&=0\\ \hat{i}(k_{3}-3)-\hat{j}(-4+k_{1})+\hat{k}(k_{2}-5)&=0\\ k_{3}-3&=0\\ 4-k_{1}&=0\\ k_{2}-5&=0\\ k_{1}&=4\\ k_{2}&=5\\ k_{3}&=3 \end{aligned}
Question 2 |
The figures show diagrammatic representations of vector fields \vec{X},\vec{Y} and \vec{Z} respectively.
Which one of the following choices is true?
\bigtriangledown \cdot \vec{X}=0, \bigtriangledown \times \vec{Y}\neq 0,\bigtriangledown \times \vec{Z}=0 | |
\bigtriangledown \cdot \vec{X} \neq 0, \bigtriangledown \times \vec{Y}=0, \bigtriangledown \times \vec{Z} \neq 0 | |
\bigtriangledown \cdot \vec{X}\neq 0, \bigtriangledown \times \vec{Y}\neq 0, \bigtriangledown \times \vec{Z}\neq 0 | |
\bigtriangledown \cdot \vec{X}=0, \bigtriangledown \times \vec{Y}= 0, \bigtriangledown \times \vec{Z}=0 |
Question 2 Explanation:
\vec{X} is going away so \vec{\triangledown } \cdot \vec{X}\neq 0
\vec{Y} is moving circulator direction so \vec{\triangledown } \cdot \vec{Y}\neq 0
\vec{Z} has circular rotation so \vec{\triangledown } \cdot \vec{Z}\neq 0
\vec{Y} is moving circulator direction so \vec{\triangledown } \cdot \vec{Y}\neq 0
\vec{Z} has circular rotation so \vec{\triangledown } \cdot \vec{Z}\neq 0
Question 3 |
The line integral of the vector field F = 5xz \hat{i}+ (3x^{2} + 2y) \hat{j} + x^{2}z\hat{k} along a path from (0,0,0) to (1,1,1) parametrized by (t, t^{2}, t) is _____.
4.41 | |
2.26 | |
6.56 | |
8.34 |
Question 3 Explanation:
\begin{aligned} E &=5xZ\bar{i}+(3x^2 +2y)\bar{j}+x^2z\bar{k} \\ &=\int _C \bar{F}\bar{d}r \\ &= \int _C 5xzdx +(3x^2+2y)dy +x^2zdz\\ x&=t, \; y=t^2,\; z=t,\; t=0 \; to \; 1 \\ dx&=dt \\ dy &=2tdt,\; dz =dt\\ &=\int_{0}^{1} 5t^2dt+(3t^2+2t^2)2tdt+t^3dt\\ &= \int_{0}^{1} (5t^2+11t^3)dt\\ &=\left [ \frac{5t^3}{3}+\frac{11t^4}{4} \right ]_0^1=\frac{5}{3}+\frac{11}{4}=4.41 \end{aligned}
Question 4 |
In cylindrical coordinate system, the potential produced by a uniform ring charge is given by \psi = f(r, z) , where f is a continuous function of r and z. Let \vec{E} be the resulting electric field. Then the magnitude of \bigtriangledown \times \vec{E}
increases with r. | |
is 0. | |
is 3. | |
decreases with z. |
Question 4 Explanation:
V is given as static field in time invariant.
Hence, \triangledown \times E=0
Hence, \triangledown \times E=0
Question 5 |
Match the following.
P-2 Q-1 R-4 S-3 | |
P-4 Q-1 R-3 S-2 | |
P-4 Q-3 R-1 S-2 | |
P-3 Q-4 R-2 S-1 |
Question 5 Explanation:
Stokes theorem \oint \vec{A}\cdot dl=\int \int (\triangledown \times A)\cdot \hat{n}ds
Gauss theorem \int \int D\cdot ds=Q
Divergence theorem \oint \oint A\cdot \hat{n}ds=\int \int \int \triangledown \cdot \bar{A}dV
Cauchy integral theorem \oint _cf(z)dz=0
Gauss theorem \int \int D\cdot ds=Q
Divergence theorem \oint \oint A\cdot \hat{n}ds=\int \int \int \triangledown \cdot \bar{A}dV
Cauchy integral theorem \oint _cf(z)dz=0
Question 6 |
Consider a function \vec{f}=\frac{1}{r^{2}}\hat{r}, where r is the distance from the origin and \hat{r} is the unit vector in the radial direction. The divergence of this function over a sphere of radius R, which includes the origin, is
0 | |
2\pi | |
4\pi | |
R\pi |
Question 6 Explanation:
\bar{f}=\frac{1}{r^2}\hat{r}
From divergence theorem as we know,
\begin{aligned} \int _{vol.} (\triangledown \cdot \bar{f})dV &=\oint \oint _S \bar{f}\cdot d\bar{S}\\ \oint \oint _S \bar{f}\cdot d\bar{S} &= \oint \oint _S \left ( \frac{1}{r^2} \cdot \hat{r}\right ) \times r^2 \sin \theta \cdot d\theta \cdot d\phi \cdot \hat{r} \\ &= \oint \oint _S \sin \theta \cdot d\theta \cdot d\phi=4\pi \end{aligned}
From divergence theorem as we know,
\begin{aligned} \int _{vol.} (\triangledown \cdot \bar{f})dV &=\oint \oint _S \bar{f}\cdot d\bar{S}\\ \oint \oint _S \bar{f}\cdot d\bar{S} &= \oint \oint _S \left ( \frac{1}{r^2} \cdot \hat{r}\right ) \times r^2 \sin \theta \cdot d\theta \cdot d\phi \cdot \hat{r} \\ &= \oint \oint _S \sin \theta \cdot d\theta \cdot d\phi=4\pi \end{aligned}
Question 7 |
The direction of vector A is radially outward from the origin, with |A|=kr^{n}, where r^{2}=x^{2}+y^{2}+z^{2} and k is a constant. The value of n for which \bigtriangledown \cdot A= 0 is
-2 | |
2 | |
1 | |
0 |
Question 7 Explanation:
\begin{aligned} \vec{A} &=kr^n\hat{i}_r \\ \triangledown \cdot \vec{A} &= \frac{1}{r^2}\frac{\partial }{\partial r}(r^2 k r^n)\\ &=\frac{1}{r^2}\frac{\partial }{\partial r}( k r^{n+2}) \\ &= (n+2)k\frac{r^{n+1}}{r^2}\\ &= (n+2)kr^{n-1}\\ \text{For}\;\; \triangledown \cdot \vec{A} &= 0\\ n+2 &=0\\ \Rightarrow \;\; n=-2 \end{aligned}
Question 8 |
Divergence of the vector field
V(x,y,z)=-(x \cos xy+y)\hat{i}+(y \cos xy)\hat{j} +(\sin z^{2}+x^{2}+y^{2})\hat{k} is
V(x,y,z)=-(x \cos xy+y)\hat{i}+(y \cos xy)\hat{j} +(\sin z^{2}+x^{2}+y^{2})\hat{k} is
2z \cos z^{2} | |
\sin xy+2z \cos z^{2} | |
x sinxy - cos z | |
None of these |
Question 8 Explanation:
\begin{aligned} V(x,y,z) &=-(x \cos xy+y)i +(y \cos xy)j \\ & \times[\sin (z^2)+(x^2)+(y^2)]k \\ \text{Divergence}&=\triangledown \cdot V \\ &=\frac{\partial V_x}{\partial x} +\frac{\partial V_y}{\partial y}+\frac{\partial V_z}{\partial z}\\ &=-\cos xy+x(\sin xy)y +\cos xy \\ &-y \sin (xy)x+2z \cos z^2 \\ &= 2z \cos z^2 \end{aligned}
Question 9 |
Consider the following statements with reference to the equation \frac{\delta p}{\delta t}
(1) This is a point form of the continuity equation.
(2) Divergence of current density is equal to the decrease of charge per unit volume per unit at every point.
(3) This is Max well's divergence equation
(4) This represents the conservation of charge
Select the correct answer.
(1) This is a point form of the continuity equation.
(2) Divergence of current density is equal to the decrease of charge per unit volume per unit at every point.
(3) This is Max well's divergence equation
(4) This represents the conservation of charge
Select the correct answer.
Only 2 and 4 are true | |
1, 2 and 3 are true | |
2, 3 and 4 are true | |
1, 2 and 4 are true |
Question 10 |
If \vec{E} is the electric intensity, \triangledown \cdot (\triangledown \times \vec{E}) is equal to
\vec{E} | |
|\vec{E}| | |
null vector | |
Zero |
Question 10 Explanation:
Divergence of a curl field is always zero.
i.e. \triangledown \cdot (\triangledown \times \vec{E})=0
i.e. \triangledown \cdot (\triangledown \times \vec{E})=0
There are 10 questions to complete.