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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:

\left [ Given \; \; \epsilon _{0} = 8.85\times 10^{-12}\: \: \: C^{2}/N-m^{2},R_{E}= 6.37\times 10^{6}m\right ]

  • 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

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

and 

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)

    CE

  • Option 2)

    CE\frac{r_{1}}{\left ( r_{2}+r \right )}

  • Option 3)

    CE\frac{r_{2}}{\left ( r+r_{2} \right )}

  • Option 4)

    CE\frac{r_{1}}{\left ( r_{1} +r\right )}

 

As we discussed in concept

Charging Of Capacitors

where in

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

at steady state there is no current through r1

 

Q=CE\frac{r_{2}}{\left ( r+r_{2} \right )}

 

 


Option 1)

CE

Option 2)

CE\frac{r_{1}}{\left ( r_{2}+r \right )}

Option 3)

CE\frac{r_{2}}{\left ( r+r_{2} \right )}

Option 4)

CE\frac{r_{1}}{\left ( r_{1} +r\right )}

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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 \epsilon_{1} at one plate to \epsilon _{2} at the other. The capacitance of capacitor is :    

 

  • Option 1)

    \epsilon _{0}\left ( \epsilon _{1} +\epsilon _{2}\right )A/d

  • Option 2)

    \epsilon _{0}\left ( \epsilon _{2} +\epsilon _{1}\right )A/2d

  • Option 3)

    \epsilon _{0}A/\left [ d\: ln\left ( \epsilon _{2}/\epsilon _{1} \right ) \right ]

  • Option 4)

    \epsilon _{0}\left ( \epsilon _{2} - \epsilon _{1}\right )A/\left [ d\: ln\left ( \epsilon _{2}/\epsilon _{1} \right ) \right ]

 

As we discussed in

If K filled between the plates -

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

 

 

- wherein

C\propto A

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)

\epsilon _{0}\left ( \epsilon _{1} +\epsilon _{2}\right )A/d

Option 2)

\epsilon _{0}\left ( \epsilon _{2} +\epsilon _{1}\right )A/2d

Option 3)

\epsilon _{0}A/\left [ d\: ln\left ( \epsilon _{2}/\epsilon _{1} \right ) \right ]

Option 4)

\epsilon _{0}\left ( \epsilon _{2} - \epsilon _{1}\right )A/\left [ d\: ln\left ( \epsilon _{2}/\epsilon _{1} \right ) \right ]

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Engineering
85 Views   |  

The  electric field in a region of space is  given by,   \vec{E}= E_{0}\: \hat{i}+2 E_{0}\: \hat{j} 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} -

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

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 :

 

\rho (r)=\rho _{0}(1-\frac{r}{R}) for  r< R

\rho (r)=0\; \; \; \; \; for\; r\geq R

Where r is the distance from the centre of the charge distribution and \rho _{0} is a constant. The electric field at an internal point (r < R) is :

  • Option 1)

    \frac{\rho _{0}}{4\epsilon _{0}}\left ( \frac{r}{3} -\frac{r^{2}}{4R}\right )

  • Option 2)

    \frac{\rho _{0}}{\epsilon _{0}}\left ( \frac{r}{3} -\frac{r^{2}}{4R}\right )

  • Option 3)

    \frac{\rho _{0}}{3\epsilon _{0}}\left ( \frac{r}{3} -\frac{r^{2}}{4R}\right )

  • Option 4)

    \frac{\rho _{0}}{12\epsilon _{0}}\left ( \frac{r}{3} -\frac{r^{2}}{4R}\right )

 

As we discussed in comcept

If P lies inside -

E_{in}=\frac{1}{4\pi \epsilon _{0}}\frac{Qr}{R^{3}}   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}}        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<R)

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)

\frac{\rho _{0}}{4\epsilon _{0}}\left ( \frac{r}{3} -\frac{r^{2}}{4R}\right )

Option 2)

\frac{\rho _{0}}{\epsilon _{0}}\left ( \frac{r}{3} -\frac{r^{2}}{4R}\right )

Option 3)

\frac{\rho _{0}}{3\epsilon _{0}}\left ( \frac{r}{3} -\frac{r^{2}}{4R}\right )

Option 4)

\frac{\rho _{0}}{12\epsilon _{0}}\left ( \frac{r}{3} -\frac{r^{2}}{4R}\right )

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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  F=-\frac{\partial U}{\partial x} 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)

    \frac{Q^{2}d}{2wl^{2}\epsilon _{0}}K

  • Option 2)

    \frac{Q^{2}w}{2dl^{2}\epsilon _{0}}(K-1)

  • Option 3)

    \frac{Q^{2}d}{2wl^{2}\epsilon _{0}}(K-1)

  • Option 4)

    \frac{Q^{2}w}{2dl^{2}\epsilon _{0}}K

 

As we discussed in concept

If K filled between the plates -

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

 

 

- wherein

C\propto A

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)

\frac{Q^{2}d}{2wl^{2}\epsilon _{0}}K

Option 2)

\frac{Q^{2}w}{2dl^{2}\epsilon _{0}}(K-1)

Option 3)

\frac{Q^{2}d}{2wl^{2}\epsilon _{0}}(K-1)

Option 4)

\frac{Q^{2}w}{2dl^{2}\epsilon _{0}}K

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Engineering
120 Views   |  

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

  • Option 1)

    \frac{1}{2}EhR

  • Option 2)

    EhR

  • Option 3)

    2EhR

  • Option 4)

    4EhR

 

As we discussed in the concept

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

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

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)

\frac{1}{2}EhR

Option 2)

EhR

Option 3)

2EhR

Option 4)

4EhR

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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  \rho . 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  \frac{qp}{3\varepsilon _{0}}

Statement 2 :  The electric field at a distance r(r < R) from the centre of the sphere is \frac{\rho r}{3\varepsilon _{0}}

 

  • 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 -

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

i.e         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 -

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 x=-a\; and\; x=a on the x- axis. A particle of mass m and charge q_{0}=\frac{q}{2} is placed at the origin. If charge q_{0} is given a small displacement (y<<a)\; along\; the\; y - axis,the net force acting on the particle is proportional to :

 

  • Option 1)

    -\frac{1}{y}

  • Option 2)

    y

  • Option 3)

    -y

  • Option 4)

    \frac{1}{y}

 

As we discussed in

Magnitude of the Resultant force -

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)

-\frac{1}{y}

Option 2)

y

Option 3)

-y

Option 4)

\frac{1}{y}

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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)

    \frac{QIn2}{4\pi \epsilon _{0}L}

  • Option 2)

    \frac{Q}{8\pi \epsilon _{0}L}

  • Option 3)

    \frac{3Q}{4\pi \epsilon _{0}L}

  • Option 4)

    \frac{Q}{4\pi \epsilon _{0}LIn2}

 

As we discussed in

Potential Difference -

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)

\frac{QIn2}{4\pi \epsilon _{0}L}

Option 2)

\frac{Q}{8\pi \epsilon _{0}L}

Option 3)

\frac{3Q}{4\pi \epsilon _{0}L}

Option 4)

\frac{Q}{4\pi \epsilon _{0}LIn2}

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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 3\times104 V/m, the charge density of the positive plate will be close to :        

 

  • Option 1)

    6\times 10^{-7} C/m^{2}

  • Option 2)

    3\times 10^{-7} C/m^{2}

  • Option 3)

    3\times 10^{4} C/m^{2}

  • Option 4)

    6\times 10^{4} C/m^{2}

 

As we discussed in the concept

Infinite Plane Parallel sheets of charge -

If  

 \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)

6\times 10^{-7} C/m^{2}

Option 2)

3\times 10^{-7} C/m^{2}

Option 3)

3\times 10^{4} C/m^{2}

Option 4)

6\times 10^{4} C/m^{2}

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Engineering
81 Views   |  

Assume that an electric field \vec{E} = 30 x^{2} \hat{i} 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 -

E_{x}=\frac{-dv}{dx}  ,  E_{y}=\frac{-dv}{dy}    ,  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 :        

K(x)=K_{0}+\lambda x(\lambda =a\: constant)

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

 

                           

 

  • Option 1)

    C=\frac{\lambda d}{ln\left ( 1+K_{0} \lambda d\right )}C_{0}

  • Option 2)

    C=\frac{\lambda }{d.ln\left ( 1+K_{0} \lambda d\right )}C_{0}

  • Option 3)

    C=\frac{\lambda d }{ln\left ( 1+ \lambda d/K_{0}\right )}C_{0}

  • Option 4)

    C=\frac{\lambda }{d.ln\left ( 1+ K_{0}/\lambda d\right )}C_{0}

 

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 -

Q\propto V

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}

C=\frac{\lambda d }{ln\left ( 1+ \lambda d/K_{0}\right )}C_{0}

 

 


Option 1)

C=\frac{\lambda d}{ln\left ( 1+K_{0} \lambda d\right )}C_{0}

Option 2)

C=\frac{\lambda }{d.ln\left ( 1+K_{0} \lambda d\right )}C_{0}

Option 3)

C=\frac{\lambda d }{ln\left ( 1+ \lambda d/K_{0}\right )}C_{0}

Option 4)

C=\frac{\lambda }{d.ln\left ( 1+ K_{0}/\lambda d\right )}C_{0}

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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 ast\rightarrow \infty is :

  • Option 1)

    3h

  • Option 2)

    \infty

  • Option 3)

    \frac{5}{3}h

  • Option 4)

    \frac{8}{3}h

 

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 \underset{p}{\rightarrow} , which makes angle θ with respect to x-axis.  When subjected to an electric field   \underset{E_{1}}{\rightarrow}  = E\hat{i} it experiences a torque    \underset{T_{1}}{\rightarrow} = \tau \hat{k}  When subjected to another electric field  \underset{E_{2}}{\rightarrow}= \sqrt{3}E_{1}\hat{j} it experiences a torque \underset{T_{2}}{\rightarrow} = \: - \underset{T_{1}}{\rightarrow}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 ~0^{\circ} and dip of θ such that   \sin \Theta = \frac{2}{3}    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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