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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

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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
Engineering
112 Views   |  

 In amplitude modulation, sinusoidal carrier frequency used is denoted by ωc and the signal frequency is denoted by ωm. The bandwidth (ωm) of the signal is such that \Delta \omega _{m}< < \omega _{m}.  Which of the following frequencies is not contained in the modulated wave ?

  • Option 1)

    ωm

  • Option 2)

     ωc

  • Option 3)

     ωmc

  • Option 4)

    ωc−ωm

 
As we have learned Side band frequency - AM wave contains three frequency   - wherein is carrier frequency, are side band frequency.    Modulated wave has frequency    Option 1) ωm Option 2)  ωc Option 3)  ωm+ωc Option 4) ωc−ωm
Engineering
285 Views   |  

The electric field component of a monochromatic radiation is given by


\underset{E}{\rightarrow} = 2E_{0}\: \: \hat{i}\: \: \cos kz\: \cos wt

Its magnetic field   \underset{B}{\rightarrow}is then given by :

 

  • Option 1)

    \frac{2E_{0}}{C}\: \hat{J}\sin kz\cos wt

  • Option 2)

    -\frac{2E_{0}}{c}\: \hat{j}\sin kz\sin wt

  • Option 3)

    \frac{2E_{0}}{c}\: \hat{j}\sin kz\sin wt

  • Option 4)

    \frac{2E_{0}}{c}\: \hat{j}\cos kz \cos wt

 
As we have learned Relation between Eo and Bo - - wherein = Electric field amplitude = Magnetic field amplitude C= Speed of light in vacuum     Electromagnetic Wave - Combination of mutually perpendicular electric and magnetic field is referred to as Electromagnetic Wave. -     It will be along Y                    Option 1) Option 2) Option 3) Option 4)
Engineering
122 Views   |  

Consider an electromagnetic wave propagating in vacuum.  Choose the correct statement :

  • Option 1)

    For an electromagnetic wave propagating in +x direction the electric field is

     \vec{E} = \frac{1}{\sqrt{2}}\, E_{yz}(x,t)\, (\hat{y}-\hat{z})    and the magnetic field is

    \vec{B} = \frac{1}{\sqrt{2}}\, B_{yz}(x,t)\, (\hat{y}+\hat{z})

     

  • Option 2)

    For an electromagnetic wave propagating in +x direction the electric field is

    \vec{E} = \frac{1}{\sqrt{2}}\, E_{yz}(y,z,t)\, (\hat{y}+\hat{z})  and the magnetic field is

    \vec{B} = \frac{1}{\sqrt{2}}\, B_{yz}(y,z,t,)\, (\hat{y}+\hat{z})

  • Option 3)

    For an electromagnetic wave propagating in +y direction the electric field is

    \vec{E} = \frac{1}{\sqrt{2}}\, E_{yz}(x,t)\, \hat{y}   and the magnetic field is

    \vec{B} = \frac{1}{\sqrt{2}}\, B_{yz}(x,t)\, \hat{z}

  • Option 4)

    For an electromagnetic wave propagating in +y direction the electric field is

    \vec{E} = \frac{1}{\sqrt{2}}\, E_{yz}(x,t)\, \hat{z}      and the magnetic field is

    \vec{B} = \frac{1}{\sqrt{2}}\, B_{z}(x,t)\, \hat{y}

 
As we have learned Electromagnetic Wave - Combination of mutually perpendicular electric and magnetic field is referred to as Electromagnetic Wave. -    If wave is propogating in x direction   must be function of (x, t )  and must be in  Y - Z plane    Option 1) For an electromagnetic wave propagating in +x direction the electric field is      and the magnetic field is   Option 2) For an...
Engineering
100 Views   |  

A conducting metal circular-wire-loop of radius r is placed perpendicular to a magnetic field which varies with time as
 B = B_{0}e^{-t/\tau } ,  where B0 and  \tau are constants, at time t = 0.  If the resistance of the loop is R then the heat generated in the loop after a long time \left ( t\rightarrow \infty \right ) is :

  • Option 1)

    \frac{\pi ^{2}r^{4}B_{0}^{4}}{2\tau R}

  • Option 2)

    \frac{\pi ^{2}r^{4}B_{0}^{2}}{2\tau R}

  • Option 3)

    \frac{\pi ^{2}r^{4}B_{0}^{2} R}{\tau }

  • Option 4)

    \frac{\pi ^{2}r^{4}B_{0}^{2} }{\tau R }

 
As we have learned Induced Current -   - wherein Resistance  Rate of change of flux         Heat generated =                                    Option 1) Option 2) Option 3) Option 4)
Engineering
91 Views   |  

 In a coil of resistance 100 , a current is induced by changing the magnetic flux through it as shown in the figure.  The magnitude of change in flux through the coil is :

  • Option 1)

    200 Wb

     

  • Option 2)

     225 Wb

     

  • Option 3)

     250 Wb

     

  • Option 4)

    275 Wb

 
As we have learned Induced Charge -   - wherein Induced Charge time independent                         Option 1) 200 Wb   Option 2)  225 Wb   Option 3)  250 Wb   Option 4) 275 Wb
Engineering
112 Views   |  

   A coil of circular cross-section having 1000 turns and 4 cm2 face area is placed with its axis parallel to a magnetic field which decreases by 10-2  Wb m-2 in 0.01 s. The e.m.f. induced in the coil is :

  • Option 1)

    400 mV

  • Option 2)

    200 mV

  • Option 3)

    4 mV

  • Option 4)

    0.4 mV

 
As we discussed in concept IF Magnetic field (B) — change - -                 Option 1) 400 mV Option 2) 200 mV Option 3) 4 mV Option 4) 0.4 mV
Engineering
98 Views   |  

In an LCR circuit as shown below both switches are open initially. Now switch S1 is closed, S2 kept open.(q is charge on the capacitor and \tau = RC is Capacitive time constant). Which of the following statement is correct ?

  • Option 1)

    At\; t=\frac{\tau }{2},q=CV(1-e^{-1})

  • Option 2)

    Work done by the battery is half of the energy dissipated in the resistor.

     

  • Option 3)

    At\; t=\tau ,q=CV/2

  • Option 4)

    At\; t=2\tau ,q=CV(1-e^{-2})

 
Charge on capacitor at t = t is    Option 1) Option 2) Work done by the battery is half of the energy dissipated in the resistor.   Option 3) Option 4)
Engineering
93 Views   |  

A diode detector is used to detect an amplitude modulated wave of 60% modulation by using a condenser of capacity 250 pico farad in parallel with a load resistance 100 kilo ohm. Find the maximum modulated frequency which could be detected by it.

 

 

  • Option 1)

    5.31 kHz

  • Option 2)

    10.62 MHz

  • Option 3)

    10.62 kHz

  • Option 4)

    5.31 MHz

 
As we have learned Modulation - It is the process carried out at transmitter in which the low frequency message signal is superimposed on a high frequency carrier signal. - wherein    We know that              Option 1) 5.31 kHz Option 2) 10.62 MHz Option 3) 10.62 kHz Option 4) 5.31 MHz
Engineering
94 Views   |  

Given : A and B are input terminals.

Logic 1 = > 5 V

Logic 0 = < 1 V

Which logic gate operation, the following circuit does ?

 

  • Option 1)

     AND Gate
     

     

  • Option 2)

     OR Gate 

  • Option 3)

    XOR Gate

     

  • Option 4)

    NOR Gate

As we have learned AND Gate - - wherein A and B are input Y is out put    When both inputs  > 5 V  then  When one of the inputs is > 5 V  and other < 1 V we get  When both input are  < 1V Then  Hence               A               B               output                            1               1                     1                           1               0                     0          ...
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