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 Multiple Choice QuestionsMultiple Choice Questions

1.

Two charges each of equal magnitude 3.2 x 10-19 coulomb but of opposite sign form an electric dipole. The distance between the two charges is 2.4 A. If the dipole is placed in an electric field of 5 x 105 volt/metre, then in equilibrium its potential energy will be

  • 3 × 15-23 joule

  • -3.84 × 10-23 joule

  • -6 × 10-23 joule 

  • -2 × 10-26 joule 


B.

-3.84 × 10-23 joule

For an electric dipole to be in stable equilibrium, we have 

Potential energy U = -pE

                           = -2qlE

                           = -q × 2l × E

Here: q = 3.2 × 10-19 coulomb, 2l = 2.4 A

            = 2.4 × 10-10 m

         E = 5 × 105 volt/metre

     U = 3.2 × 10-19 × 2.4 × 10-10 × 5×105 J

            = -3.84 × 10-23 J


2.

Length cannot be measure by

  • fermi

  • micron

  • debye

  • light year


C.

debye

A fermi is equivalent to a femtometer and it is older non-SI measurement unit of length.

A micron is a unit of measure in the system equal to 1 millionth of a meter in length ( about 39 millionths of an inch ).

The light-year is a unit of length used to express astronomical distances and measures about 9.46 trillion kilometers ( 9.46 × 1012 km )

Debye is the unit of electric dipole moment. Therefore length cannot be measured by debye.


3.

Two spheres of same metal have radii a and b. Thet have been connected to a conducting wire. Find the ratio of the electric field intensity upon them.

  • a/b

  • b/a

  • b2 / a

  • b2/a2


D.

b2/a2

Since both the metal spheres are connected to the same conducting wire, both of them will be having same charge on them.

Let the charge on each of them by q

Then the electric field intensities are given by

         Eakqa2

         Ebkqb2

∴         EaEb = b2a2


4.

Figure shows the electric lines of forces emerging from a charged body. If the electric field at A and B are EA and  EB respectively and if the displacement between A and B is r, then

    

  • EA > EB

  • EA < EB

  • EA = EB/r

  • EA = E/ r2


A.

EA > EB

Since the intensity of the electric lines of force at A is more than that at B, the electric field at A.  EA  >  EB ( electric field at B ).


5.

Using mass (M), length (L), time (T ) and current (A) as fundamental quantities, the dimension of permeability is

  • [ M-1 L T-2 A ]

  • [ M L2 T-2 A-1 ]

  • [ M L T-2A-2 ]

  • [ M L T-1 A-1 ]


C.

[ M L T-2A-2 ]

We know that the force per unit length of a wire carrying current due to another parallel wire increase of temperature. Which in turn implies that carrying current is given by

   dFdt  =  μoi1i22πd

⇒  μo2πdi1 i2 dFdl

   [ μo ] = LA2 · M L T-2 L

    [ μo ] = [ M L T-2 ] [A-2 ]


6.

A small piece of metal wire is dragged across the gap between the pole pieces of a magnet in 0.4 sec. If magnetic flux between the pole pieces is known to be 8 × 10-4 Wb, then induced emf in the wire, is

  • 4 × 10-3 V

  • 8 × 10-3 V 

  • 20 × 10-3 V

  • 6 × 10-3 V


C.

20 × 10-3 V

The charge of flux through the metal wire 

      Δϕ = 8 × 10-4 Wb

Time taken  

    Δt = 0.4 sec

∴  Induced e.m.f

     e =  ϕt

        =  8 × 10-40.4

  e = 20 × 10-3 V


7.

1 newton/coulomb is equivalent to

  • 1 C/V

  • 1 J

  • 1 V/M

  • 1 J/C


C.

1 V/M

From the relation,

Force F = qE

   Fq = E = Vd

   newtoncoulomb = voltmetre


8.

Qεo

  • Q6εo

  • Q6 εo

  • Qεo a2

  • Q4πεoa2


B.

Q6 εo

Gauss's law tells that the total flux through an arras enclosing a charge Q is  Qεo.

Now as the cube is having six faces and as we can assume a symmetrical distribution of fluxes among its faces, the flux associated with one of its face is  Q6εo.


9.

The unit of permittivity (ε0) of space is

  • newton-metre/ coulomb2

  • coulomb / newton-metre

  • coulomb / newton-metre2

  • coulomb/ newton-metre2


D.

coulomb/ newton-metre2

From the relation

          F = 14πε0q1q2r2

or  4πε01F q1q2r2 

 Unit of ε0 = coulomb2newton metre2

                   =  coulomb/ newton metre2


10.

An electron is moving with velocity v on a circular path of radius r in a transverse electric field B. The specific charge (e/m) of the electron is

  • Bvr

  • νBr

  • Bvr2

  • B


B.

νBr

If a magnetic field is applied perpendicular to a moving electron, then electron will move on a circular path. 

 centripetal force = force on electron due to magnetic field

           2r = Beν

            em = vrB