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 Multiple Choice QuestionsMultiple Choice Questions

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1.

The plates of a parallel plate capacitor of capacity of 50 µF are charged by a battery to a potential of 100 volt. The battery remains connected the plates are separated from each other so that the distance between them is doubled. Then, the energy spent by battery in doing so, will be

  • 12.5×10-2J

  • 35×10-2J

  • 42.5×10-2J

  • 25×10-2J


D.

25×10-2J

When the separation between the plates is doubled the capacitance becomes one half.

It means C'=50/2=25μF

Now energy spent by the battery =qV=(C'V)V

=C'V2=25×10-6×(100)2

=25×10-2J


2.

From the figure find the capacitance of the capacitor?


B.

The given capacitor can be viewed as a series combination of two capacitors C1 and C2 where 

where d is the seperation between the plates

       

∴ The effective capacitance


3.

 The capacity of a parallel plate capacitor with no dielectric substance but with a separation of 0.4 cm is 2µF. The separation is reduced to half and it is filled with a dielectric substance of value of 2.8. The final capacity of capacitor is

  • 15.6μF

  • 11.2μF

  • 19μF

  • 22.4μF


B.

11.2μF

C=ε0 Ad1dHence, C1C2=d2d1 or2C2=0.20.4or       C2=0.4×20.2=4μFFinal capacitor isC =4×k=4×2.8=11.2μF 


4.

If a unit positive charge is taken from one point to another over an equipotential surface, then

  • work is done on the charge

  • work is done by the charge

  • work done is constant

  • no work is done


D.

no work is done

Since the potential at each point of an equipotential surface is the same. the potential does not change while we move a unit positive charge from one point to another. Therefore work done in the process is zero.


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5.

Two small magnets each of magnetic moment 10 Am2 are placed in end on position 0.1 m apart from their centres. The force acting between them is

  • 0.6N

  • 0.06N

  • 0.06×107N

  • 0.6×107N


A.

0.6N

F=μo4π6M1M2r4=10-7×6×10×100.14 =0.6N


6.

An electron is travelling along the x-direction. It encounters a magnetic field in the y-direction. Its subsequent motion will be

  • straight line along the x-direction

  • a circle in the xz plane

  • a circle in the yz plane

  • a circle in the xy plane


B.

a circle in the xz plane

We know that the force on a charged particle e due to a magnetic field is given by

    F = e  v × B 

Here v = vx  = i^  and  B  = By j^

∴    F = e vx By i × j

       F = evx By k^

Therefore the subsequent motion of charged particle will be a circle in the xz plane.


7.

Two metal plates having a potential difference of 800 V and 2 cm apart. It is found that a particle of mass 1.96 x 10-15 kg remaining suspended in the region between the plates. The charge on the particle must be (e = elementry charge)

  • 8e

  • 6e

  • 3e

  • 2e


C.

3e

E=Vd=8002×10-2Vm=4×104 Vmagain, qE =mgq=mgE=1.96×10-15×9.84×104e   q=3e


8.

An electron having charge 'e' and mass m is moving in a uniform electric field E. Its acceleration will be

  • e2m

  • eEm

  • eE2m

  • mEe


B.

eEm

Charge on electron  = e

Mass of electron = m

intensity of uniform electric field = E

Acceleration of the electron

        α = Force on the electronMass of electron

            = Electric charge × electric fieldMass of electron

       α = eEm


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9.

In the given figure, the capacitance C1,  C3,  C4 and  C5 have  a capacitance 4 μF each. If the capacitor  C2 has a capacitance 10 μF, then effective capacitance between A and B will be

    

  • 2 μF

  • 6 μF

  • 4 μF

  • 8 μF


C.

4 μF

Capacitance of capacitors   C1 , C3 ,  C4 , C5  = 4 μF each and capacitance of capacitor C2 = 10 μF.

If a battery is applied across A and B, the points b and c will be at the same potential ( since C1 = C4 = C3 = C5 = 4 μF )

Therefore no charge flows through C2.

We have the capacitors C1 and C5 in series.

Therefore their equivalent capacitance

  C'' = C3 × C5C1 + C5

       = 4 × 44 + 4

  C" = 2 μF

Now C' and C" are in parallel. Therefore capacitance between A and B 

      = C' + C"

      = 2 + 2

      = 4 μF


10.

The working of dynamic is based on the principle of

  • Electromagnetic induction

  • chemical effect of current

  • magnetic effect of current

  • heating effect of current


A.

Electromagnetic induction

The dynamo operates on the principle of the production of dynamically induced emf. Hence whenever flux is cut by the conductor, emf is produced in it according to the law of electromagnetic induction.


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