Question 1 |
Let a_r, a_\phi \; and \; a_z be unit vectors along r, \phi \; and \; z directions, respectively in the cylindrical
coordinate system. For the electric flux density given by D = (a_r15 + a_\phi 2r - a_z3rz) Coulomb/m^2, the total electric flux, in Coulomb, emanating from the volume enclosed
by a solid cylinder of radius 3 m and height 5 m oriented along the z-axis with its base
at the origin is:
108 \pi | |
54 \pi | |
90 \pi | |
180 \pi |
Question 1 Explanation:
\begin{aligned}
& \psi / \text{Crossing closed surface} \\ \oint \oint \vec{D}.\vec{ds}&=\int \int \int (\vec{\triangledown }\cdot \vec{D}) dv \\ (\vec{V}.\vec{D}) &=\frac{1}{\rho }\frac{\partial }{\partial p}(\rho D_{\rho })+\frac{1}{\rho }\frac{\partial D_{\phi }}{\partial \phi }+\frac{\partial D_{z}}{\partial z} \\ &=\frac{1}{\rho }\frac{\partial }{\partial p}(\rho 15)+\frac{1}{\rho }\frac{\partial }{\partial \phi }(2\rho )+\frac{\partial }{\partial z}(-3\rho z) \\ &=\frac{1}{\rho }15-3\rho \\ \int \int \int (\vec{\triangledown }\cdot \vec{D})dv&=\int \int \int \left ( \frac{15}{\rho}-3\rho \right )\rho d\rho d\phi dz \\ &=\int \int \int 15d\rho d\phi dz-3\int \int \int \rho ^{2}d\rho d\phi dz \\ &=15\int_{\rho =0}^{3}d\rho \int_{\phi =0}^{2\pi }d\phi \int_{z=0}^{5}-3\int_{\rho =0}^{3}\rho^{2}d\rho \int_{\phi =0}^{2\pi }d\phi \int_{z=0}^{5}dz \\ &= 15(3-0)(2\pi )(5)-3\left ( \frac{3^{3}}{3} \right )\times (2\pi )(5) \\ &=45(10\pi )-27(10\pi )=180\pi C
\end{aligned}
Question 2 |
The static electric field inside a dielectric medium with relative permittivity, \varepsilon _r = 2.25,
expressed in cylindrical coordinate system is given by the following expression
E=a_r 2r+a_\varphi \left ( \frac{3}{r} \right )+a_z6
where a_r,a_\varphi ,a_z are unit vectors along r, \varphi \; and \;z directions, respectively. If the above expression represents a valid electrostatic field inside the medium, then the volume charge density associated with this field in terms of free space permittivity, \varepsilon _0, in SI units is given by:
E=a_r 2r+a_\varphi \left ( \frac{3}{r} \right )+a_z6
where a_r,a_\varphi ,a_z are unit vectors along r, \varphi \; and \;z directions, respectively. If the above expression represents a valid electrostatic field inside the medium, then the volume charge density associated with this field in terms of free space permittivity, \varepsilon _0, in SI units is given by:
3\varepsilon _0 | |
4\varepsilon _0 | |
5\varepsilon _0 | |
9\varepsilon _0 |
Question 2 Explanation:
\vec{D}=\epsilon \vec{E}=\epsilon _{o}\epsilon _{r}\vec{E}
\vec{D}=\epsilon _{o}2.25\left ( 2r\hat{a_{r}}+\frac{3}{r}\hat{a_{\phi}+6\hat{a_{z}}} \right )
\vec{D}=4.5\epsilon _{o}r\hat{a_{r}}+\frac{6.75\epsilon _{o}}{r}\hat{a_{\phi }}+13.5\epsilon _{o}\hat{a_{z}}
volume charge density
\rho _{v}=\vec{\bigtriangledown }\cdot \vec{D}
\rho _{v}=\frac{1}{r}\frac{\partial }{\partial r}(rD_{r})+\frac{1}{r}\frac{\partial D_{\phi }}{\partial \phi }+\frac{\partial D_{z}}{\partial z}
\rho _{v}=\frac{1}{r}\frac{\partial }{\partial r}(r4.5\epsilon _{o}r)+\frac{1}{r}\frac{\partial }{\partial \phi }\left ( \frac{6.75\epsilon _{o}}{r} \right )+\frac{\partial }{\partial z}(13.5\epsilon _{o})
\, \, \, =\frac{1}{r}\frac{\partial }{\partial r}(4.5\epsilon _{o}r^{2})+0+0
\, \, \, =\frac{1}{r}(4.5\epsilon _{o})(2r)=9\epsilon _{o}
\vec{D}=\epsilon _{o}2.25\left ( 2r\hat{a_{r}}+\frac{3}{r}\hat{a_{\phi}+6\hat{a_{z}}} \right )
\vec{D}=4.5\epsilon _{o}r\hat{a_{r}}+\frac{6.75\epsilon _{o}}{r}\hat{a_{\phi }}+13.5\epsilon _{o}\hat{a_{z}}
volume charge density
\rho _{v}=\vec{\bigtriangledown }\cdot \vec{D}
\rho _{v}=\frac{1}{r}\frac{\partial }{\partial r}(rD_{r})+\frac{1}{r}\frac{\partial D_{\phi }}{\partial \phi }+\frac{\partial D_{z}}{\partial z}
\rho _{v}=\frac{1}{r}\frac{\partial }{\partial r}(r4.5\epsilon _{o}r)+\frac{1}{r}\frac{\partial }{\partial \phi }\left ( \frac{6.75\epsilon _{o}}{r} \right )+\frac{\partial }{\partial z}(13.5\epsilon _{o})
\, \, \, =\frac{1}{r}\frac{\partial }{\partial r}(4.5\epsilon _{o}r^{2})+0+0
\, \, \, =\frac{1}{r}(4.5\epsilon _{o})(2r)=9\epsilon _{o}
Question 3 |
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 3 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 4 |
A co-axial cylindrical capacitor shown in Figure (i) has dielectric with relative permittivity \varepsilon _{r1}=2. When one-fourth portion of the dielectric is replaced with another dielectric ofrelative permittivity \varepsilon _{r2}, as shown in Figure (ii), the capacitance is doubled. The value of \varepsilon _{r12} is ____.
10 | |
7 | |
15 | |
18 |
Question 4 Explanation:
Co-axial cylindrical capacitor-1
\begin{aligned} C_1 &=\frac{2 \pi \in _h}{\ln\left ( \frac{b}{a} \right )}=\frac{2 \pi \in _0 (2)h}{\ln \left ( \frac{R}{r} \right )}\\ C_1&=\frac{2 \pi \in _0 h}{\ln \left ( \frac{R}{r} \right )}\;\;...(i) \end{aligned}
Co-axial cylindrical capacitor-2
\begin{aligned} C_2 &=\frac{\left (\frac{3 \pi}{2} \right ) \in _0 (2)h}{\ln \left ( \frac{R}{r} \right )}+\frac{\frac{\pi}{2} \in _0 \in _{r_2}h}{\ln \left ( \frac{R}{r} \right )}\\ C_2&=\frac{ \pi \in _0 h}{\ln \left ( \frac{R}{r} \right )}\left [ 3+\frac{\in _{r_2}}{2} \right ]\;\;...(ii) \end{aligned}
Given, C_2=2C_1 ....(iii)
Put equation (i), (ii) in equation (iii),
\begin{aligned} &\frac{ \pi \in _0 h}{\ln \left ( \frac{R}{r} \right )}\left [ 3+\frac{\in _{r_2}}{2} \right ]=2\left [ \frac{4 \pi \in _0 h}{\ln \left ( \frac{R}{r} \right )} \right ]\\ &3+\frac{\in {r_2}}{2}=8\\ &\Rightarrow \; \frac{\in {r_2}}{2}=5\; \Rightarrow \; \in {r_2}=10 \end{aligned}
\begin{aligned} C_1 &=\frac{2 \pi \in _h}{\ln\left ( \frac{b}{a} \right )}=\frac{2 \pi \in _0 (2)h}{\ln \left ( \frac{R}{r} \right )}\\ C_1&=\frac{2 \pi \in _0 h}{\ln \left ( \frac{R}{r} \right )}\;\;...(i) \end{aligned}
Co-axial cylindrical capacitor-2
\begin{aligned} C_2 &=\frac{\left (\frac{3 \pi}{2} \right ) \in _0 (2)h}{\ln \left ( \frac{R}{r} \right )}+\frac{\frac{\pi}{2} \in _0 \in _{r_2}h}{\ln \left ( \frac{R}{r} \right )}\\ C_2&=\frac{ \pi \in _0 h}{\ln \left ( \frac{R}{r} \right )}\left [ 3+\frac{\in _{r_2}}{2} \right ]\;\;...(ii) \end{aligned}
Given, C_2=2C_1 ....(iii)
Put equation (i), (ii) in equation (iii),
\begin{aligned} &\frac{ \pi \in _0 h}{\ln \left ( \frac{R}{r} \right )}\left [ 3+\frac{\in _{r_2}}{2} \right ]=2\left [ \frac{4 \pi \in _0 h}{\ln \left ( \frac{R}{r} \right )} \right ]\\ &3+\frac{\in {r_2}}{2}=8\\ &\Rightarrow \; \frac{\in {r_2}}{2}=5\; \Rightarrow \; \in {r_2}=10 \end{aligned}
Question 5 |
The capacitance of an air-filled parallel-plate capacitor is 60 pF. When a dielectric slab
whose thickness is half the distance between the plates, is placed on one of the plates
covering it entirely, the capacitance becomes 86 pF.Neglecting the fringing effects, the
relative permittivity of the dielectric is _____________ (up to 2 decimal places).
1.55 | |
2.53 | |
3.75 | |
4.25 |
Question 5 Explanation:
Given, C=\frac{\varepsilon _0A}{d}=60pF
In second case:
Capacitance,
\begin{aligned} C_1 &=\frac{\varepsilon _0A}{d/2} =\frac{2\varepsilon _0A}{d}\\ &= 2 \times (60 pF)=120pF\\ C_2&=\frac{2\varepsilon _0\varepsilon _rA}{d} \\ &=(2 \times 60)\varepsilon _r \;pF =120\varepsilon _r \; pF\\ C_{eq}&=\frac{C_1C_2}{C_1+C_2}=\frac{120 \times 120\varepsilon _r}{(120+120\varepsilon _r)}pF \\ &= 86\;pF \;\;(given)\\ 86&=\frac{120 \times 120\varepsilon _r}{120(1+\varepsilon _r)} \\ \frac{86}{120}&= \frac{\varepsilon _r}{1+\varepsilon _r}\\ \varepsilon _r &=\frac{86}{34}=2.53 \end{aligned}
In second case:
Capacitance,
\begin{aligned} C_1 &=\frac{\varepsilon _0A}{d/2} =\frac{2\varepsilon _0A}{d}\\ &= 2 \times (60 pF)=120pF\\ C_2&=\frac{2\varepsilon _0\varepsilon _rA}{d} \\ &=(2 \times 60)\varepsilon _r \;pF =120\varepsilon _r \; pF\\ C_{eq}&=\frac{C_1C_2}{C_1+C_2}=\frac{120 \times 120\varepsilon _r}{(120+120\varepsilon _r)}pF \\ &= 86\;pF \;\;(given)\\ 86&=\frac{120 \times 120\varepsilon _r}{120(1+\varepsilon _r)} \\ \frac{86}{120}&= \frac{\varepsilon _r}{1+\varepsilon _r}\\ \varepsilon _r &=\frac{86}{34}=2.53 \end{aligned}
Question 6 |
A positive charge of 1 nC is placed at (0,0,0.2) where all dimensions are in metres.
Consider the x-y plane to be a conducting ground plane. Take \epsilon _{0}=8.85 \times 10^{-12} F/m. The z component of the E field at (0,0,0.1) is closest to
899.18 V/m | |
-899.18 V/m | |
999.09 V/m | |
-999.09 V/m |
Question 6 Explanation:
Net electric field at point P due to charge Q is,
\begin{aligned} \vec{E_{12}}&=\frac{Q\vec{R_{12}}}{4 \pi \varepsilon _0|\vec{R_{12}}|^3} \\ &= \frac{1 \times 10^{-9}(-0.1\hat{a_z})}{4 \pi (8.854 \times 10^{-12})(0.1)^3}\\ &+ \frac{-1 \times 10^{-9}(-0.3\hat{a_z})}{4 \pi (8.854 \times 10^{-12})(0.3)^3}\\ &= \left [ \frac{-10^5}{4 \pi (8.854)}- \frac{10^5}{4 \pi (8.854)9} \right ]\hat{a_z}\\ &=(-898.774-99.863) \hat{a_z}\\ &= -999.09\hat{a_z}\; V/m \end{aligned}
Question 7 |
A thin soap bubble of radius R = 1 cm, and thickness a=3.3\mu m(a\lt \lt R), is at a potential of 1 V with respect to a reference point at infinity. The bubble bursts and becomes a single spherical
drop of soap (assuming all the soap is contained in the drop) of radius r. The volume of the soap
in the thin bubble is 4\pi R^{2}a and that of the drop is \frac{4}{3} \pi r^{3}. The potential in volts, of the resulting single spherical drop with respect to the same reference point at infinity is __________. (Give the answer up to two decimal places.)
10.03 | |
20.25 | |
30.28 | |
40.08 |
Question 7 Explanation:
After burst,
4 \pi R^2 a=\frac{4}{3}\pi r^3
Radius of soap drop =(3 R^2a)^{1/3}=0.099665cm
Initial voltage was 1 V and C=4 \pi \varepsilon _0R
and initial charge, Q=(4 \pi \varepsilon _0R \times 1)
Since after bursting Q remainsame and C=4 \pi \varepsilon _0r
New potential on soap drop,
V=\frac{Q}{C}=\frac{4 \pi \varepsilon _0R}{4 \pi \varepsilon _0r}=\frac{1}{0.099665}=10.03V
4 \pi R^2 a=\frac{4}{3}\pi r^3
Radius of soap drop =(3 R^2a)^{1/3}=0.099665cm
Initial voltage was 1 V and C=4 \pi \varepsilon _0R
and initial charge, Q=(4 \pi \varepsilon _0R \times 1)
Since after bursting Q remainsame and C=4 \pi \varepsilon _0r
New potential on soap drop,
V=\frac{Q}{C}=\frac{4 \pi \varepsilon _0R}{4 \pi \varepsilon _0r}=\frac{1}{0.099665}=10.03V
Question 8 |
Consider a solid sphere of radius 5 cm made of a perfect electric conductor. If one million
electrons are added to this sphere, these electrons will be distributed.
uniformly over the entire volume of the sphere | |
uniformly over the outer surface of the sphere | |
concentrated around the centre of the sphere | |
along a straight line passing through the centre of the sphere |
Question 8 Explanation:
Added charge (one million electrons) to be solid spherical conductor is uniformly distributed over the outer surface of the sphere.
Question 9 |
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 9 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 10 |
The magnitude of magnetic flux density (B) in micro Teslas (\muT) at the center of a loop of wire wound as a regular hexagon of side length 1m carrying a current (I=1A), and placed in vacuum as shown in the figure is __________.
1.4 | |
2.8 | |
0.7 | |
0.2 |
Question 10 Explanation:
i= 1 A
Here, B at point P is
B=\frac{\mu _0I}{4 \pi d}(\sin \theta _1 +\sin \theta _2)
We know for each segment of hexagon
magnetic field intensity due to element length dx,
dH=\frac{Idx}{4 \pi r^2}i \times u_r
where,
i \times u_r=(\sin \theta)\hat{k}....perpendicular to plane of paper
\begin{aligned} H &=\int dH=\int \frac{I \sin \theta}{4 \pi r^2}(dx)\hat{k} \\ B&=\mu _0 H \\ B_p&=\frac{\mu _0I}{4 \pi d}(\cos \alpha _1+\cos \alpha _2)\hat{k} \\ d&=\frac{\sqrt{3}}{2} l\;\;\;...\text{where}\; l=1m\\ \alpha _1&=\alpha _2 =60^{\circ}\\ |B_p| &=\frac{\mu _0 I}{4 \pi \frac{\sqrt{3}}{2} }(\cos 60^{\circ} +\cos 60^{\circ}) \\ &\text{for all six segment} \\ 6 \times |B_p| &=\frac{4 \pi \times 10^{-7} \times 6 \times 1}{4 \pi \times \frac{\sqrt{3}}{2} }(1) \\ &=\frac{12}{\sqrt{3}}\times 10^{-7}=\frac{1.20}{\sqrt{3}}\times 10^{-6}T \\ B&=0.69 \mu T \end{aligned}
Here, B at point P is
B=\frac{\mu _0I}{4 \pi d}(\sin \theta _1 +\sin \theta _2)
We know for each segment of hexagon
magnetic field intensity due to element length dx,
dH=\frac{Idx}{4 \pi r^2}i \times u_r
where,
i \times u_r=(\sin \theta)\hat{k}....perpendicular to plane of paper
\begin{aligned} H &=\int dH=\int \frac{I \sin \theta}{4 \pi r^2}(dx)\hat{k} \\ B&=\mu _0 H \\ B_p&=\frac{\mu _0I}{4 \pi d}(\cos \alpha _1+\cos \alpha _2)\hat{k} \\ d&=\frac{\sqrt{3}}{2} l\;\;\;...\text{where}\; l=1m\\ \alpha _1&=\alpha _2 =60^{\circ}\\ |B_p| &=\frac{\mu _0 I}{4 \pi \frac{\sqrt{3}}{2} }(\cos 60^{\circ} +\cos 60^{\circ}) \\ &\text{for all six segment} \\ 6 \times |B_p| &=\frac{4 \pi \times 10^{-7} \times 6 \times 1}{4 \pi \times \frac{\sqrt{3}}{2} }(1) \\ &=\frac{12}{\sqrt{3}}\times 10^{-7}=\frac{1.20}{\sqrt{3}}\times 10^{-6}T \\ B&=0.69 \mu T \end{aligned}
There are 10 questions to complete.