Q&A - Ask Doubts and Get Answers

Sort by :
Clear All
Q
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}

View More
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}

View More
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

View More
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}

View More
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...
Engineering
296 Views   |  

Figure shows a circular area of radius R where a uniform magnetic field \vec{B} is going into the plane of paper and increasing in magnitude at a constant rate. In that case, which of the following graphs, drawn schematically, correctly shows the variation of the induced electric field E(r) ?

  • Option 1)

  • Option 2)

  • Option 3)

  • Option 4)

 
As we discussed in concept Induced electric field - When A constant B Varying   - wherein                         Option 1) Option 2) Option 3) Option 4)
Engineering
137 Views   |  


 A 2V battery is connected across AB as shown in the figure. The value of the current supplied by the battery when in one case battery’s positive terminal is connected to A and in other case when positive terminal of battery is connected to B will respectively be :

 

  • Option 1)

    0.2 A and 0.1 A
     

  • Option 2)

    0.4 A and 0.2 A

  • Option 3)

    0.1 A and 0.2 A

     

  • Option 4)

     0.2 A and 0.4 A

 
As we have learned P -N junction as diode - It is a one way device. It offers a low resistance when forward biased and high resistance when reverse biased. - wherein R = 0, Forward  R  Reverse     In one case  NO current will pass through  I = 2/5 A = 0.4 A  In second case : no current will pass through  I = 2/10 = 0.2 A      Option 1) 0.2 A and 0.1 A   Option 2) 0.4 A and 0.2 A Option...
Engineering
382 Views   |  

 Three straight parallel current carrying  conductors are shown in the figure. The force experienced by the middle conductor of length 25 cm is :

  • Option 1)

       3 x 10-4  N  toward right

  • Option 2)

       6 x 10-4  N  toward left

  • Option 3)

       9 x 10-4  N  toward left

  • Option 4)

    Zero

 
As we have learned Force between two parallel current carrying conductors - - wherein I1 and I2 current carrying two parallel wires  a-seperation between two wires      Force due to wire I    towards right  Force due to wire two    towards left  Net force =                Option 1)    3 x 10-4  N  toward right Option 2)    6 x 10-4  N  toward left Option 3)    9 x 10-4  N  toward...
Engineering
125 Views   |  

 The circuit shown here has two batteries of 8.0 V and 16.0 V and three resistors 3 \Omega, 9 \Omega and 9 \Omega and a capacitor 5.0 \muF.

How much is the current I in the circuit in steady state ?

  • Option 1)

     1.6 A
     

     

  • Option 2)

     0.67 A

  • Option 3)

     2.5 A

     

  • Option 4)

     0.25 A

 

As we have learned

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

 

 

I'_1 = I_2 = I  since in steady state current through capacitor is  0 

USing kirchoff's rule 

16-9I -3I -8V = 0   or I = 8/12 A = 2/3 A 

 = 0.67 A 

 

 


Option 1)

 1.6 A
 

 

Option 2)

 0.67 A

Option 3)

 2.5 A

 

Option 4)

 0.25 A

View More
Engineering
289 Views   |  

Two identical wires A and B, each of length ‘l’, carry the same current I.  Wire A is bent into a circle of radius R and wire B is bent to form a square of side ‘a’.  If BA and BB are the values of magnetic field at the centres of the circle and square
respectively, then the ratio\frac{B_{A}}{B_{B}} is

  • Option 1)

    \frac{\pi ^{2}}{ }

  • Option 2)

    \frac{\pi ^{2}}{16\sqrt{2}}

  • Option 3)

    \frac{\pi ^{2}}{16}

  • Option 4)

    \frac{\pi ^{2}}{8\sqrt{2}}

 
As we have learnt,   Magnetic Field due to circular coil at Centre - - wherein X = 0    Magnetic field at the centre of circle                            in case of square Option 1) Option 2) Option 3) Option 4)
Engineering
508 Views   |  

The box of a pin hole camera, of length L, has a hole of radius a.  It is assumed that when the hole is illuminated by a parallel beam of light of wavelength λ the spread of the spot (obtained on the opposite wall of the camera) is the sum of its geometrical spread and the spread due to diffraction. The spot would then have its minimum size (say bmin) when :

  • Option 1)

    a= \frac{\lambda ^{2}}{L}\: \: \: and\: \: b_{min}= \left ( \frac{2\lambda ^{2}}{L} \right )

  • Option 2)

    a= \sqrt{\lambda L}\: \: \: and\: \: b_{min}= \left ( \frac{2\lambda ^{2}}{L} \right )

  • Option 3)

    a= \sqrt{\lambda L}\: \: \: and\: \: b_{min}= \sqrt{4\lambda L}

  • Option 4)

    a=\frac{\lambda ^{2}}{L}\: \: and \: \: b_{min}=\sqrt{4\lambda L}

 
As we have learnt,   Fraunhofer Diffraction -   - wherein Condition of nth minima. slit width angle of deviation     Considering the gemetrical spreading of light alone O'A' = OA = Q Freinel's displacement  Considering the Freinel's class of diffraction Angle of diffraction is given by the relation From fig:  From figure     Option 1) Option 2) Option 3) Option 4)
Engineering
277 Views   |  

In the following, which one of the diodes is reverse  biased?

  • Option 1)

      

  • Option 2)

  • Option 3)

  • Option 4)

 
As we have learnt,   P -N junction as diode - It is a one way device. It offers a low resistance when forward biased and high resistance when reverse biased. - wherein R = 0, Forward  R  Reverse    Correct option is 1 In all other figure, diode is forward biased.     Option 1)    Option 2) Option 3) Option 4)
Engineering
110 Views   |  

A particle is released on a vertical smooth semicircular track from point X so that OX makes angle \Theta from the vertical (see figure). The normal reaction of the track on the particle vanishes at point Y where OY makes angle \Phi with the horizontal. Then :

  • Option 1)

    \sin \Phi =\cos \Theta

  • Option 2)

    \sin \Phi = \frac{1}{2}\cos \Theta

  • Option 3)

    \sin \Phi = \frac{2}{3}\cos \Theta

  • Option 4)

    \sin \Phi = \frac{3}{4}\cos \Theta

 
As we have learned If only conservative forces act on a system, total mechnical energy remains constant - -     let velocity at point Y  is   From energy conservatiuon  At Y            Option 1) Option 2) Option 3) Option 4)
Engineering
100 Views   |  

The graph between angle of deviation (\delta ) and angle of incidence (i) for a triangular prism is represented by :

 

 

 

  • Option 1)

  • Option 2)

  • Option 3)

  • Option 4)

 
As we have learned Condition of minimum deviation -     - wherein       Option 1) Option 2) Option 3) Option 4)
Engineering
93 Views   |  

If a white light is used in Young’s double slit experiment, then a very large number of coloured fringes can be seen:

  • Option 1)

    With first order violet fringes being closer to the central white fringes

  • Option 2)

    First order red fringes being closer to the central white fringes

  • Option 3)

    With a central white fringe

  • Option 4)

    With a central black fringe

 
As we have learned @2539 When white light is used the central fringe is white with coloured fringes around that fringe  Option 1) With first order violet fringes being closer to the central white fringes Option 2) First order red fringes being closer to the central white fringes Option 3) With a central white fringe Option 4) With a central black fringe
Engineering
95 Views   |  

A converging lens is used to form an image on a screen. When the upper half of the lens is covered by an opaque screen:

  • Option 1)

    half of the image will disappear

  • Option 2)

    image will not form on the screen

  • Option 3)

    intensity of image will increase

  • Option 4)

    intensity of image will decrease

 
  As we have learned @2468WH When half f the lrns is converged by opaque screen , only intensity reduced but no effect on formation of image  Option 1) half of the image will disappear Option 2) image will not form on the screen Option 3) intensity of image will increase Option 4) intensity of image will decrease
Exams
Articles
Questions