Question # 1

The minimum charge on any object can not be less than

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1.6 x 10^-19C

3.2 x 10^-19C

1.0 C

4.8 x 10^-19C

A point charge Q is placed at the mid-point of a line joining two charges. 4q and q. if the net force on charge q is zero. then Q must be equal to

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-q

+q

-2q

+4q

10⁶electrons are moving through a wire per second, the current developed is

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1.6 x 10^-19

1 A

1.6 x 10^-15A

10⁶A

The force of repulsion between two point charges is F, when these are at a distance 0.1 m apart. Now the point charges are replaced by sphere of radii 5 cm each having the same charge as that of the respective point charges. The distance between their centre is again kept 0.1 m ; then the force of repulsion will

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Increase

Decrease

Remain F

Become 10F/9

A closed surface contains two equal and opposite charges. The net electric flux from the surface will be

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Negative

Positive

Infinite

Zero

A wire of radius r has resistance R. If it is stretched to a wire of r/2 radius, then the resistance becomes

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2R

4R

16R

Zero

Capacitance of two or more capacitors

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Increases in series combination

Increases in parallel combination

Remains unchanged

None of the above

A parallel plate capacitor is first charged and then a dielectric slab is introduced between the plates. The quantity that remains unchanged is

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Charge Q

Potential V

Capacity

Energy U

The electric intensity outside the two oppositely charged parallel metal plates is

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Maximum

Minimum

Zero

Infinite

If the distance of separation between two chares is increased, the electrical potential energy of the system will

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Increase

Decrease

May increase or decrease

Remain the same

The SI unit of electric field intensity is

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CN^-1

NC^-1or Vm^-1

JC^-1

AV^-1

A heater coil rated at (1000 W - 200 V) is connected to 110 volt line. What will be the power consumed?

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200 W

302.5

250 W

350 W

A current of 1.6 A is passed through a solution of CuSO_{4. How many Cu}²⁺ions are liberated in one minute?

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3 x 10²⁰

3 x 10¹⁰

6 x 10²⁰

6 x 10¹⁰

The electric potential at the surface of an atomic nucleus (Z = 50) of radius 9.0 x 10^-15is

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9 x 10⁵V

9 V

8 x 10⁶V

80 V

Cause of heat production in a current carrying conductor is

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Collisions of free electrons with one another

High drift speed of free electrons

Collisions of free electrons with atoms or ions of conductor

High resistance value

A metal plate of thickness half the separation between the capacitor plates of capacitance C is inserted. The new capacitance is

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C

C/2

Zero

2C

The unit of intensity of electric field is

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newton/coluomb

jule/coluomb

volt x metre

newton/metre

If the two charges in Coulomb's law have double distance between them, then electric force

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Becomes two-fold

Becomes four-fold

Remains the same

None of these

The capacitance of a parallel plate capacitor depends upon

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Area of the plates

Separation between the plates

Medium between the plates

All of the above

The minimum resistance that can be obtained by connecting 5 resistance of 1/4Ωeach is

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4/5Ω

5/4Ω

20Ω

0.05 Ω

Thermocouple is an arrangement of two different metals

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To convert heat energy in to electrical energy

To produce more heat

To convert heat energy into chemical energy

To convert electric energy in to heat energy

The unit of resistivity is

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ohm

ohm-m²

ohm-meter

ohm-m^-1

The capacity of a parallel plat capacitor depends on the

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Type to metal used

Thickness of plates

Potential applied across the plates

Separation between the plates

Electric generators which convert mechanical energy into

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solar energy

thermal energy

kinetic energy

electrical energy

The conductivity of a superconductor is

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Infinite

Very large

Very small

Zero

The energy required to charge a capacitor of 5μF by connecting D.C. source of 20 KV is

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10 KJ

5 KJ

2 KJ

1 KJ

Force acting upon a charged particle kept between the plates of a charged condenser if F. IF one of the plates of the condenser is removed, force acting on the same will become

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Zero

F/2

F

2F

An electric dipole is at the centre of a hollow sphere of radius r. The total normal electric flux through the sphere is (here Q is the charge and d is the distance between the two charges of the dipole)

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Q/4πr²

2Q/4πr²

Q.d

Zero

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5μF

10μF

3μF

6μF

The relation between charge 'Q' and current 'I' is given by

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Q = I/t

Q = It

Q = I²t

Q = I²/t

The ratio of the gravitational force F_gto the electrostatic force F_ebetween two electrons at the same distance apart is approximately

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9.8

24 x 10¹⁹

24 x 10⁴²

24 x 10^-44

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