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Engineering
96 Views   |

The magnitude of the average electric field normally present in the atmosphere just above the surface of the Earth is about

150 N/C, directed inward towards the center of the Earth.  This gives the total net surface charge carried by the Earth to  be:

• Option 1)

+670 kC

• Option 2)

- 670 kC

• Option 3)

- 680 kC

• Option 4)

+ 680 kC

As we discussed in the concept

Infinite Plane parallel sheets of charge -

If

$\dpi{100} \sigma _{A}=\sigma$ and $\dpi{100} \sigma _{B}=-\sigma$ $\dpi{100} \rightarrow$  $\dpi{100} E_{p}= E_{R}=0$

and

$\dpi{100} E_{Q}= \frac{\sigma }{\epsilon _{0}}$

-

Electric Field E = 150 N/C

Total surface charge carried by earth q= ?

$q=\varepsilon _{0EA} = \varepsilon _{0}E \pi r^{2}$

$= 8.85\times 10^{-12}\times 150\times (6.37\times 10^{6})^{2}$

$\simeq 680 KC$

As electric field is directed inwards, hence, $q=-680 KC$

Option 1)

+670 kC

Option 2)

- 670 kC

Option 3)

- 680 kC

Option 4)

+ 680 kC

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Engineering
194 Views   |

In the given circuit diagram when the current reaches steady state in the circuit, the charge on the capacitor of capacitance C will be :

• Option 1)

• Option 2)

• Option 3)

• Option 4)

As we discussed in concept

where in

# $\dpi{100} Q=Q_{0}\left ( 1-e^{\frac{-t}{Rc}} \right )$

at steady state there is no current through r1

Q=

Option 1)

Option 2)

Option 3)

Option 4)

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Engineering
89 Views   |

The gap between the plates of a parallel plate capacitor of area A and distance between plates d, is filled with a dielectric whose permittivity varies linearly from at one plate to at the other. The capacitance of capacitor is :

• Option 1)

• Option 2)

• Option 3)

• Option 4)

As we discussed in

If K filled between the plates -

$\dpi{100} {C}'=K\frac{\epsilon _{0}A}{d}={C}'=Ck$

- wherein

$\dpi{100} C\propto A$

$\dpi{100} C\propto\frac{1}{d}$

$dV=\frac{E_0}{k}dx\\V=\int_{0}^{d}\frac{\sigma dx}{\varepsilon _0\frac{(\varepsilon _2-\varepsilon _1)}{d}x+\varepsilon _1}\\V=\frac{\varepsilon d}{\varepsilon _0(\varepsilon _2-\varepsilon _1)}ln\frac{\varepsilon _2}{\varepsilon _1}\\Q=CV\\ \epsilon _{0}\left ( \epsilon _{2} - \epsilon _{1}\right )A/\left [ d\: ln\left ( \epsilon _{2}/\epsilon _{1} \right ) \right ]$

Option 1)

Option 2)

Option 3)

Option 4)

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Engineering
85 Views   |

The  electric field in a region of space is  given by,    where E0=100 N/C. The flux of this field through a circular surface of radius 0.02 m parallel to the Y-Z plane is nearly :

• Option 1)

0.125 Nm2/C

• Option 2)

0.02 Nm2/C

• Option 3)

0.005 Nm2/C

• Option 4)

3.14 Nm2/C

As we discussed in

Electric field \vec{E} through any area \vec{A} -

$\dpi{100} \phi = \vec{E}\cdot \vec{A}=EA\cos \Theta$

$\dpi{100} S.I\; unit\; -\left ( volt \right )m\; or\; \frac{N-m^{2}}{c}$

- wherein

$\underset{E}{\rightarrow}= E_0\hat{i} + 2E_0\hat{J}$

$E_0= 100W/C$

$\vec{E}= 100\hat{i} + 200 \hat{J}$

$A= \pi r^{2} = \frac{22}{7}\times 0.02\times 0.02$

$A= 1.25\times 10^{-3}\hat{i}m^{2}$

$\therefore$ New flux $\therefore\phi =E A cos\theta$

$\phi =(100\hat{i} + 200\hat{J}). 1.25\times 10^{-3}\hat{i}cos\theta$

where $\theta = 0$

$\phi = 1.25\times 10^{-3} Nm^{2}/c$

$= 0.125 Nm^{2}/C$

Option 1)

0.125 Nm2/C

Option 2)

0.02 Nm2/C

Option 3)

0.005 Nm2/C

Option 4)

3.14 Nm2/C

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Engineering
129 Views   |

A spherically symmetric charge  distribution is characterised by a charge density having the following variation :

for  r< R

Where r is the distance from the centre of the charge distribution and is a constant. The electric field at an internal point (r < R) is :

• Option 1)

• Option 2)

• Option 3)

• Option 4)

As we discussed in comcept

If P lies inside -

$\dpi{100} E_{in}=\frac{1}{4\pi \epsilon _{0}}\frac{Qr}{R^{3}}$   $\dpi{100} V_{in}=\frac{Q}{4\pi \epsilon _{0}}\frac{3R^{2}-r^{2}}{2R^{3}}$

$\dpi{100} E_{in}=\frac{\rho r}{3 \epsilon _{0}}$        $\dpi{100} V_{in}=\frac{\rho \left ( 3R^{2}-r^{2} \right )}{6 \epsilon _{0}}$

-

$dq=\int 4 \pi x^{2} dx=\rho _{0} (1- \frac{x}{R})4 \pi x^{2} dx$

where $(x

$q=\int dq= 4\pi \rho _{0}\int_{0}^{r}(1-\frac{x}{R})x^{2}dx$

=$4\pi\rho _{0}[\frac{x^{2}}{3}-\frac{x^{4}}{4R}]^{r}$

=$4\pi\rho _{0}[\frac{r^{2}}{3}-\frac{r^{4}}{4R}]$

$q= 4\pi\rho _{0}r^{3}[\frac{1}{3}-\frac{r}{4R}]$

$E=\frac{1}{4 \pi \varepsilon _{0}}\frac{q}{r^{2}}= \frac{1}{4 \pi \varepsilon _{0}}\frac{4 \pi\ \varepsilon _{0}r^{3}}{r^{2}}[\frac{1}{3}-\frac{r}{4R}]$

$E=\frac{\rho _{0}}{\varepsilon ^{0}}[\frac{r}{3}-\frac{r^{2}}{4R}]$

Option 1)

Option 2)

Option 3)

Option 4)

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Engineering
144 Views   |

A parallel plate capacitor is made of two plates of length l, width w and separated by distance d. A dielectric slab (dielectric constant K) that fits exactly between the plates is held near the edge of the plates. It is pulled into the capacitor by a force  where U is the energy of the capacitor when dielectric is inside the capacitor up to distance x (See figure). If the charge on the capacitor is Q then the force on the dielectric when it is near the edge is :

• Option 1)

• Option 2)

• Option 3)

• Option 4)

As we discussed in concept

If K filled between the plates -

$\dpi{100} {C}'=K\frac{\epsilon _{0}A}{d}={C}'=Ck$

- wherein

$\dpi{100} C\propto A$

$\dpi{100} C\propto\frac{1}{d}$

$C= C_{1}+ C_{2}= \frac{K(x\omega )\varepsilon _{0}}{d} + \frac{(l-x)\omega \varepsilon _{0}}{d}$

$C=\frac{\omega \varepsilon _{0}}{d}\times (Kx + (l-x))$

$v=\frac{1}{2}\times \frac{Q^{2}}{C}= \frac{Q^{2}d}{2\omega \varepsilon _{0}(\varepsilon +(k-1)x)}$

$\frac{\partial v }{\partial x}=-\frac{dQ^{2}(K-1)}{2\omega \varepsilon _{0}(l+(k-1)x)^{2}}$

$F=-\frac{\partial v}{\partial x}= \frac{Q^{2}d(K-1)}{2\omega l^{2}\varepsilon _{0}} at x=0$

Option 1)

Option 2)

Option 3)

Option 4)

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Engineering
120 Views   |

A cone of base radius R and height h is  located in a uniform electric field parallel to its base. The electric flux entering the cone is :

• Option 1)

• Option 2)

• Option 3)

• Option 4)

As we discussed in the concept

Electric field \vec{E} through any area \vec{A} -

$\dpi{100} \phi = \vec{E}\cdot \vec{A}=EA\cos \Theta$

$\dpi{100} S.I\; unit\; -\left ( volt \right )m\; or\; \frac{N-m^{2}}{c}$

- wherein

Area of $\Delta$ facing = $\frac{1}{2}\times h\times 2R$

$\therefore \phi = EhR$

Option 1)

Option 2)

Option 3)

Option 4)

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Engineering
158 Views   |

This question has statement 1 and statement 2.  Of the four choices given after the statements, choose the one that best describes the two statements.

An insulating solid sphere of radius R has a uniformly positive charge density  . As a result of this uniform charge distribution there is a finite value of electric potential at the centre of the sphere, at the surface of the sphere and also at a point out side the sphere. The electric potential at infinity is zero.

Statement 1 : When a charge q is taken from the centre to the surface of the sphere, its potential energy changes by

Statement 2 :  The electric field at a distance r(r < R) from the centre of the sphere is

• Option 1)

Statement 1 is true, Statement 2 is true, Statement 2 is not the correct explanation for statement 1.

• Option 2)

Statement 1 is true, Statement 2 is false

• Option 3)

Statement 1 is false, Statement 2 is true

• Option 4)

Statement 1 is true, Statement 2 is the correct explanation for statement 1

As we discussed in

If P lies at centre r = 0 -

$\dpi{100} V_{centre}=\frac{3}{2}\times \frac{1}{4\pi \epsilon _{0}}\frac{Q}{R}=\frac{3}{2}V_{s}$

i.e         $\dpi{100} V_{c}> V_{s}> V_{o}$

-

Potential at the centre of the sphere,

$V_{C}= \frac{R^{2}\rho }{2\varepsilon _{0}}$

Potential at the surface of the sphere,

$V_{S}= \frac{1}{3}\frac{R^{2}\rho }{\varepsilon _{0}}$

When a charge q is taken from the centre to the surface, the change in potential energy is

$\Delta U=\left ( V_{C} -V_{S}\right )q= \left (\frac{R^{2}\rho }{2\varepsilon _{0}} -\frac{1}{3}\frac{R^{2}\rho }{\varepsilon _{0}} \right )q= \frac{1}{6}\frac{R^{2}\rho q}{\varepsilon _{0}}$

Statement 1 is false. Statement 2 is true.

Option 1)

Statement 1 is true, Statement 2 is true, Statement 2 is not the correct explanation for statement 1.

Option 2)

Statement 1 is true, Statement 2 is false

Option 3)

Statement 1 is false, Statement 2 is true

Option 4)

Statement 1 is true, Statement 2 is the correct explanation for statement 1

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Engineering
93 Views   |

In a uniformly charged sphere of total charge Q and radius R the electric field E is plotted as a function of distance from the centre. The graph which would correspond to the above will be :

• Option 1)

• Option 2)

• Option 3)

• Option 4)

:    For uniformly charged sphere

As we discussed in

Graph - - wherein

The variation of  E   with distance  r   from the centre is as shown.

$E= \frac{1}{4\pi \varepsilon _{0}}\frac{Qr}{R^{3}}\: \: \left ( For\: r< R \right )$

$E= \frac{1}{4\pi \varepsilon _{0}}\frac{Q}{R^{2}}\: \: \left ( For\: r= R \right )$

$E= \frac{1}{4\pi \varepsilon _{0}}\frac{Q}{r^{2}}\: \: \left ( For\: r> R \right )$

Option 1)

Option 2)

Option 3)

Option 4)

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Engineering
147 Views   |

Two capacitors C1 and C2 are charged to 120 V and 200 V respectively. It is found that by connecting them together the potential of each one can be made zero. Then :

• Option 1)

9C1 = 4C2

• Option 2)

5C1 = 3C2

• Option 3)

3C1 = 5C2

• Option 4)

3C1 + 5C2 = 0

As we discussed in

Parallel Grouping -

$\dpi{100} C_{eq}=C_{1}+C_{2}+\cdots$

- wherein

$120C_1 = 200C_2$

$6C_1 = 10C_2$

$3C_1 = 5C_2$

Option 1)

9C1 = 4C2

Option 2)

5C1 = 3C2

Option 3)

3C1 = 5C2

Option 4)

3C1 + 5C2 = 0

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Engineering
85 Views   |

Two charges, each equal to q, are kept at on the - axis. A particle of mass m and charge is placed at the origin. If charge is given a small displacement - axis,the net force acting on the particle is proportional to :

• Option 1)

• Option 2)

• Option 3)

• Option 4)

As we discussed in

Magnitude of the Resultant force -

$\dpi{100} F_{net}=\sqrt{F_{1}^{2}+F_{2}^{2}+2F_{1}F_{2}\cos \Theta }$

- wherein

$F_{net} = 2Fcos\theta$

$F_{net} = \frac {2Kq(\frac{q}{2})Y}{(\sqrt{Y^2 + a^2})^2(\sqrt{Y^2 + a^2})}$

$F_{net} = \frac {2Kq(\frac{q}{2})Y}{{Y^2 + a^2})^\frac{3}{2}}=\frac{Kq^2Y}{a^3}$

$\therefore F \propto Y$

Option 1)

Option 2)

Option 3)

Option 4)

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Engineering
133 Views   |

A charge Q is uniformly distributed over a long rod AB of length L, as shown in the figure. The electric potential at the poing O lying at a distance L from the end A is :

• Option 1)

• Option 2)

• Option 3)

• Option 4)

As we discussed in

Potential Difference -

$\dpi{100} V_{B}-V_{A}=\frac{w}{q}$

-

Charge on the element

$dQ = \frac{Q}{L}dx$

Potential at 0

$dV = \frac{1}{4\pi\epsilon_o}\frac{dQ}{x} = \frac{1}{4\pi\epsilon_o}\frac{Q}{Lx}dx$

$\int dV = \int_{L}^{2L}\frac{1}{4\pi\epsilon_o}\frac{Q}{Lx} dx = \frac{1}{4\pi\epsilon_o}\frac{Q}{L}[lnx]_{L}^{2L}$

$V = \frac{Q ln2}{4\pi\epsilon_o L}$

Option 1)

Option 2)

Option 3)

Option 4)

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Engineering
158 Views   |

A parallel plate capacitor is made of two   circular plates separated by a distance of 5 mm and with a dielectric of dielectric constant 2.2 between them.  When the electric field in the dielectric is 3104 V/m, the charge density of the positive plate will be close to :

• Option 1)

• Option 2)

• Option 3)

• Option 4)

As we discussed in the concept

Infinite Plane Parallel sheets of charge -

If

$\dpi{100} \sigma _{A}=\sigma _{B}=\sigma \rightarrow \left | E_{p} \right |=\left | E_{R} \right |=\frac{\sigma }{\epsilon _{0}}$

and  EQ = 0

-

$E=\frac{\sigma }{K \epsilon_{0}}$

$\therefore$ Charge density $\sigma =K \epsilon _{0}E$

= $2.2\times 8.85\times 10^{-12}\times 3\times 10^{4}$

$\sigma = 6\times 10^{-7} c/m^{2}$

Option 1)

Option 2)

Option 3)

Option 4)

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Engineering
81 Views   |

Assume that an electric field exists in space.  Then the potential difference VA - VO, where VO is the potential at the origin and VA the potential at x=2 m is :

• Option 1)

120 J

• Option 2)

-120 J

• Option 3)

- 80 J

• Option 4)

80 J

As we discussed in the concept

In space -

$\dpi{100} E_{x}=\frac{-dv}{dx}$  ,  $\dpi{100} E_{y}=\frac{-dv}{dy}$    ,  $\dpi{100} E_{z}=\frac{-dv}{dz}$

-

$\underset{E}{\rightarrow}=30x^{2}\iota$

$=>dv=-\vec{E}. \vec{dx}=\int_{V_{1}}^{V_{4}}dv= -\int_{0}^{2} 30x^{2} dx$

$V_{a}- V_{0} = -10\left | 8 \right | J= -80 J$

Option 1)

120 J

Option 2)

-120 J

Option 3)

- 80 J

Option 4)

80 J

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Engineering
169 Views   |

The space between the plates of a parallel plate capacitor is filled with a ‘dielectric’  whose ‘dielectric constant’ varies with  distance as per the relation :

The capacitance C, of this capacitor, would   be related to its ‘vacuum’ capacitance Co as per the relation :

• Option 1)

• Option 2)

• Option 3)

• Option 4)

As we discussed in

The boundary Conditions -

$\dpi{100} dV=-\int_{r_{1}}^{r_{2}}\overrightarrow{E}\cdot \vec{d}r=-\int_{r_{1}}^{r_{2}}Edr\cos \theta$

-

Capacitance of Conductor -

$\dpi{100} Q\propto V$

$\dpi{100} Q=CV$

- wherein

C - Capacity or capacitance of conductor

V - Potential.

Given $K=K_{0}+\lambda x$

$V= -\int_{0}^{d} Edr= V=\int_{0}^{d}\frac{\sigma }{K\epsilon _{0}} dx$

$V= \frac{\sigma}{\varepsilon _{0}} \int_{0}^{d}\frac{1}{K+\lambda x} dx= \frac{\sigma }{\lambda\varepsilon _{0} }[ln(K_{0}+\lambda d)-lnK_{0}]$

$V= \frac{\sigma }{\lambda\varepsilon _{0} }ln (1+\frac{\lambda d}{K_0})$

$C= \frac{Q}{V}=\frac{\sigma S}{V}= \frac{\sigma S}{\frac{\sigma }{\lambda }ln(1+\frac{\lambda d}{K_0})}$ S= surface area of plate.

here, $C_0=\frac{\varepsilon _{0}S}{d}$

Option 1)

Option 2)

Option 3)

Option 4)

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Engineering
339 Views   |

An object is dropped from a height h from the ground.  Every time it hits the ground it looses 50% of its kinetic energy.  The total distance covered as is :

• Option 1)

3h

• Option 2)

• Option 3)

• Option 4)

First is right answer, because every time it hit ground it cannot  much distance

Engineering
313 Views   |

Assume that a neutron breaks into a proton and an electron. The energy released during this process is (Mass of neutron = 1.6725 x 10–27kg; mass of proton = 1.6725 x 10–27kg; mass of electron = 9 x 10–31 kg)

• Option 1)

0.73 MeV

• Option 2)

7.10 MeV

• Option 3)

6.30 MeV

• Option 4)

0.51 MeV

Mass defect, Option 1) 0.73 MeV Option 2) 7.10 MeV Option 3) 6.30 MeV Option 4) 0.51 MeV
Engineering
371 Views   |

A player caught a cricket ball of mass 150 g  moving  at a rate of 20 m/s . If the catching process is completed in 0.1 s  the force of the blow exerted by the ball on the hand of the player is equal to

• Option 1)

300 N

• Option 2)

150 N

• Option 3)

3 N

• Option 4)

30 N

As we learnt in Impulse Momentum Theorem - - wherein If   is increased, average force is decreased       Force time = Impulse = Change of momentum Correct option is 4. Option 1) 300 N This is an incorrect option. Option 2) 150 N This is an incorrect option. Option 3) 3 N This is an incorrect option. Option 4) 30 N This is the correct option.
Engineering
119 Views   |

An electric dipole has a fixed dipole moment , which makes angle θ with respect to x-axis.  When subjected to an electric field      it experiences a torque      When subjected to another electric field   it experiences a torque  The angle θ is :

• Option 1)

300

• Option 2)

450

• Option 3)

600

• Option 4)

900

As we have learned Torque Experienced by the dipole -                    - wherein       Form (1) and (2)          Option 1) 300 Option 2) 450 Option 3) 600 Option 4) 900
Engineering
323 Views   |

A fighter plane of length 20 m, wing span (distance from tip of one wing to the tip of the other wing) of 15 m and height 5 m is flying towards east over Delhi.  Its speed is 240 ms−1.  The earth’s magnetic field over Delhi is 5×10−5 T with the declination angle ~ and dip of θ such that      If the voltage developed is VB between the lower and upper side of the plane and VW between the tips of the wings then VB and VW are close to :

• Option 1)

VB = 45 mVVW = 120 mV with right side of pilot at higher voltage

• Option 2)

VB = 45 mV ; VW = 120 mV with left side of pilot at higher voltage

• Option 3)

VB = 40 mV ; VW = 135 mV with right side of pilot at higher voltage

• Option 4)

VB = 40 mV ; VW = 135 mV with left side of pilot at higher voltage

As we have learned Motional EMF -   - wherein magnetic field length velocity of u perpendicular to uniform magnetic field.                       Option 1)  VB = 45 mV ;  VW = 120 mV with right side of pilot at higher voltage Option 2) VB = 45 mV ; VW = 120 mV with left side of pilot at higher voltage Option 3) VB = 40 mV ; VW = 135 mV with right side of pilot at higher voltage Option...
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