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Engineering
74 Views   |  

The supply voltage to a room is 120 V. The resistance of the lead wires is 6 \Omega. A 60 W bulb is already switched on. What is the decrease of voltage across the bulb, when a 240 W heater is switched on in parallel to the bulb?

 

  • Option 1)

    10.04 Volt

  • Option 2)

    Zero Volt

  • Option 3)

    2.9 Volt

  • Option 4)

    13.3 Volt

 

Need details on this question

Engineering
77 Views   |  

   If \lambda _{0} and \lambda be the threshold wavelength and wavelength of incident light, the velocity of photoelectron ejected from the metal surface is :

  • Option 1)

    \sqrt{\frac{2h}{m}\left ( \lambda _{0}-\lambda \right )}

  • Option 2)

    \sqrt{\frac{2hc}{m}\left ( \lambda _{0}-\lambda \right )}

  • Option 3)

    \sqrt{\frac{2hc}{m}\left ( \frac{\lambda _{0}-\lambda }{\lambda \lambda _{0}} \right )}

  • Option 4)

    \sqrt{\frac{2h}{m}\left ( \frac{1}{\lambda _{0}}-\frac{1}{\lambda } \right )}

 

As discussed in the concept

Photoelectric Effect -

\frac{1}{2}mu^{2}= hv-hv_{0}

- wherein

where

m is the mass of the electron

u is the velocity associated with the ejected electron.

h is plank’s constant.

v is frequency of photon,

v0 is threshold frequency of metal.

 

 v^{2} = \frac{2hc}{m}\left ( \frac{1}{\lambda } - \frac{1}{\lambda _{0}} \right )

v =\sqrt{ \frac{2hc}{m}\left ( \frac{1}{\lambda } - \frac{1}{\lambda _{0}} \right )}

 

Hence the correct option is 3

 


Option 1)

\sqrt{\frac{2h}{m}\left ( \lambda _{0}-\lambda \right )}

Option 2)

\sqrt{\frac{2hc}{m}\left ( \lambda _{0}-\lambda \right )}

Option 3)

\sqrt{\frac{2hc}{m}\left ( \frac{\lambda _{0}-\lambda }{\lambda \lambda _{0}} \right )}

Option 4)

\sqrt{\frac{2h}{m}\left ( \frac{1}{\lambda _{0}}-\frac{1}{\lambda } \right )}

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Engineering
72 Views   |  

 Chloro compound of Vanadium has only spin magnetic moment of 1.73 BM.  This Vanadium chloride has the formula :

(at. no. of V=23)

  • Option 1)

    VCl2

  • Option 2)

    VCl4

  • Option 3)

    VCl3

  • Option 4)

    VCl5

 

As discussed in

Magnetic Quantum Number (m) -

It  gives information about the spatial orientation of the orbital with respect to standard set of co-ordinate axis.

-

 

 The value of magnetic moment is 1.73 BM

1.73 = \sqrt{n\left ( n+2 \right )} where n is the number of unpaired electrons

3 = n \left ( n+2 \right )

After calculation n = 1

V_{23} => 1S^{2},2S^{2},2p^{6},3S^{2},3p^{6},4S^{2},3d^{3}

To obtain one unpaired electron V should be tetrapositive ion and the formula of its chlorid should be VCl_{4}

Option 2 is correct

 

 

 

 


Option 1)

VCl2

Incorrect option 

Option 2)

VCl4

Correct option

Option 3)

VCl3

Incorrect option

Option 4)

VCl5

Incorrect option

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Engineering
63 Views   |  

Resistance of a given wire is obtained  by measuring the current flowing in it and the voltage difference applied across it. If the percentage errors in the measurement of the current and the voltage difference are 3% each, then error in the value of resistance of the wire is

  • Option 1)

    6%

  • Option 2)

    zero

  • Option 3)

    1%

  • Option 4)

    3%

 

As discussed in

Ohm's Law -

Current flowing through the conductor is directly proportional to the Potential difference accross two ends .

- wherein

unit -   \frac{m^{2}}{volt\cdot sec}

 

V \alpha  I

V=IR

R- Electric Resistance

 

 

\dpi{100} R=\frac{V}{I}

\therefore \; \; \frac{\Delta R}{R}=\frac{\Delta V}{V}+\frac{\Delta I}{I}

Percentage error in R is

\frac{\Delta R}{R}\times 100=\frac{\Delta V}{V}\times 100+\frac{\Delta I}{I}\times 100

=3\%+3\%=6\%  

 


Option 1)

6%

This option is correct.

Option 2)

zero

This option is incorrect.

Option 3)

1%

This option is incorrect.

Option 4)

3%

This option is incorrect.

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Engineering
65 Views   |  

This question has Statement I and Statement II. Of the four choices given after the Statements, choose the one that best describes the two Statemens.

Statement - I : Higher the range, greater is the resistance of ammeter.

Statement - II : To increase the range of ammeter,additional shunt needs to be used across it.

 

 

 

  • Option 1)

    Statement - I is false, Statement - II is true.

     

  • Option 2)

    Statement - I is true, Statement - II is true, Statement - II is the correct explanation of Statement - I.

     

  • Option 3)

    Statement - I is true, Statement - II is true, Statement - II is not the correct explanation of Statement - I.

     

  • Option 4)

    Statement - I is true, Statement - II is false.

     

 

As discussed in-

Required shunt -

s= \frac{i_{g}G}{(i-i_{g})}

- wherein

i_{g}- Current through galvanometer

 

 The answer follows from basic Principle of stunt and its use to make an ammeter:

s=\frac{I_{g}}{I-I_{g}}


Option 1)

Statement - I is false, Statement - II is true.

 

This option is correct.

Option 2)

Statement - I is true, Statement - II is true, Statement - II is the correct explanation of Statement - I.

 

This option is incorrect.

Option 3)

Statement - I is true, Statement - II is true, Statement - II is not the correct explanation of Statement - I.

 

This option is incorrect.

Option 4)

Statement - I is true, Statement - II is false.

 

This option is incorrect.

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Engineering
77 Views   |  

A 10 V battery with internal resistance1\Omega and a 15V battery with internal resistance 0.6\Omega are connected in parallel to a voltmeter (see figure). The reading in the voltmeter will be close to :

  • Option 1)

    11.9 V

  • Option 2)

    12.5 V

  • Option 3)

    13.1 V

  • Option 4)

    24.5 V

 

As we discussed in

In closed loop -

-i_{1}}R_{1} + i_{2}}R_{2} -E_{1}-i_{3}}R_{3}+E_{2}+E_{3}-i_{4}}R_{4}=0

- wherein

 

 Current in the circuit 

I=\frac{5}{1.6}=\frac{50}{16}=\frac{25}{8}A

Reading of the voltmeter 

V=15-\frac{25}{8}\times 0.6

V=15-\frac{15}{8}=13.1V


Option 1)

11.9 V

The option is incorrect 

Option 2)

12.5 V

The option is incorrect 

Option 3)

13.1 V

The option is correct 

Option 4)

24.5 V

The option is incorrect 

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Engineering
27 Views   |  

A 3 olt battery with  negligible internal resistance is connected in a circuit as shown in the figure. The current I i n the circuit will be

  • Option 1)

    1A

  • Option 2)

    1.5 A

  • Option 3)

    2A

  • Option 4)

    (1/3)A

 

as discussed in

Parallel Grouping -

Potential - Same

Current - Different

- wherein

 

 Req=\frac{(3+3)\times3}{(3+3)+3}=\frac{18}{9}=2\Omega

I=\frac{V}{R}=\frac{3}{2}=1.5A


Option 1)

1A

This option is incorrect.

Option 2)

1.5 A

This option is correct.

Option 3)

2A

This option is incorrect.

Option 4)

(1/3)A

This option is incorrect.

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Engineering
32 Views   |  

Two electric bulbs marked  25 W-220 V and 100 W - 220 V are connected in series to a 440 V supply .Which of the bulbs will fuse?

  • Option 1)

    both

  • Option 2)

    100W

  • Option 3)

    25 W

  • Option 4)

    neither

 

As discussed in:

Series Grouping -

Potential - Different

Current - Same

- wherein

 

I_{1}=\frac{P_{1}}{V_{1}}

I_{1}=\frac{25}{220}Amp

Similarly,

I_{2}=\frac{P_{2}}{V_{2}}=\frac{100}{220}Amp

The current flowing through the circuit I=\frac{440}{Reff}

Reff= R_{1}+R_{2}

R_{1}= \frac{V^{2}_{1}}{P_{1}}=\frac{(220)^{2}}{25}

R_{2}= \frac{V^{2}_{2}}{P_{1}}=\frac{(220)^{2}}{100}

\therefore I=\frac{440}{\frac{(220)^{2}}{25}+\frac{(220)^{2}}{100}}

After solving this, we get I=\frac{40}{220}Amp

\because I_{1}(=\frac{25}{220}A)< I(=\frac{40}{220}A)<I_{2}(=\frac{100}{200}A)

Thus the bulb rated 25W-220V will fuse. 

 


Option 1)

both

Option 2)

100W

Option 3)

25 W

Option 4)

neither

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Engineering
176 Views   |  


Three capacitances, each of 3 \muF, are  provided. These cannot be combined to provide the resultant capacitance of :

 

  • Option 1)

    1\mu F

  • Option 2)

    2\mu F

  • Option 3)

    4.5\mu F

  • Option 4)

    6\mu F

 

3

 

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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