2ND YEAR PHYSICS NET mcq, Exams of Physics

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J S A c a d e m y F a i s a l a b a d Learn to Live, Live to Learn

JSA Series Important MCQS Physics Part-II

---

NOTE Four possible answers are given against each question in columns A, B, C and D. Select the right answer and on the separate Answer Sheet , fill the

circle A,B,C or D with pen or marker in front of that question number.

CH#12(Electrostatics)

S.# QUESTIONS A B C D

Large number of

small number of

Large number of free small number of

1 Metals are good conductors of electricity because they have bounded

bounded electrons electrons

free electrons electrons

Two oppositely charged balls A & B attract third ball C when placed

Positively charged

Electrically

Negatively charged

Both (a) & (c)



near them turn by turn. Then ball C must be neutral

3 Free electrons are Tightly bound fixed Strongly fixed Loosely bound



4 SI unit of charge is calorie

com

ampere volt Coulomb



5 The number of free electrons in one coulomb charge is zero

1.6x

6.2x

20

6.2x

18 

Charles Augustus Coulomb measured the force between two charges blogspot 

If the atomic number of copper is 29, the contribution of electrons per



atom in the block of copper will be

7 Charge on an electron was determined by Ampere Maxwell Millikan



Thompson

by

TaleemTutor

Physical. balance Common balance Cavendish balance Torsion balance

9 If the distance between two charges is doubled, the electric forc F=2(kq 1 q 2 /r

2

) F=1/2(kq 1 q 2 /r

2

) F=4(kq 1 q 2 /r

2

) F=1/4(kq 1 q 2 /r

2

)



between them will become

If the distance between two charges is doubled, the electric force

Four times One half twice

One fourth



between them will become

The electric force between two charges placed in air is 2N. when



placed in a medium of εr=80, the force reduces to

12 The force in medium of relative permittivity εr is given by Fmed= εr /F Fmed=F εr F=Fmed / εr Fmed=F/ εr



13 Electric charge of 100μC is 13cm apart from another charge 16.9μC. 9x

7

N 9x

5

N 90 N 900 N



The force between them in Newton is

The force b/w two point charges in air or vacuum is F. if air or vacuum

decreases



14 is replaced by an insulator of relative permittivity εr the force b/w Remains constant Becomes infinite increases

charges

15 Value of dielectric constant for air or vacuum is Greater than one Less than one zero one



16 The electrostatic force of repulsion between two electrons at a 1.8 N 2.30 x 10

N 2.30 x 10

N (^) 2.30 x 10

N

distance of one meter is

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JSA Series Important MCQS Physics Part-II

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17 The magnitude of charge on an electron is 1.6x

10

C 1.6x

C 1.6x

19

C 1.6x

 C



18 Concept of electric field theory was introduced by Lenz Coulomb Joseph Henry

Michael

Faraday



Force experienced by a unit positive charge placed at a point in an

Capacity Electric potential Magnetic field

Electric field

electric field is known as intensity



20 The force per unit charge is known as Electric flux Electric potential Electron volt

Electric field

intensity



21 The SI unit of electric field intensity(strength) E is Nm

2

Nm

2

C

N-1^ m-2C^2 N/C



22 E = F/r

2

q/F Fq F/q



23 An electric field can deflect Neutrons Gamma rays x-rays Alpha particles



24 An electric field can not deflect Alpha particles Electrons Protons x-rays^



b/w two

Near a positive p int Near a negative near two oppositely oppositely

25 The electric field will be uniform

.

blogspot com

charge point^ charged bodies^ charged parallel

metal plates



26 Electric field intensity due to a point charge at distance r is equal to 4πєє 0 (q/r) є 0 /4πє(q

2

/r) 4πє/ є 0 (q/r

2

) q / 4πєє 0 r

2 

27 Electric flux is given by the formula EA/sinθ ExA EAcosθ E. A



intensity TaleemTutor (^). direction of the

All of above



28 Electric flux through any surface depend on ea of the surface electric field intensity surface

29 According to gauss’s law , the flux through any closed surface is Φ=1/Qεε 0 Φ= ε 0 /Qε Φ=Qεε 0 Φ=Qε/ ε 0



30 Gauss’s law can only be applied to a------------- surface Curved flat closed Any shape



Which of the following can be taken as measure of electric fie d

Φ=Qε/ A F/A Qε/ ε 0 A^

Φ/ A



32 When a surface is held parallel to E then flux is infinite maximum negative zero



33 SI unit of electric flux is N/C (^) Nm (^2) C-2 (^) N-1 (^) m-2C (^2) NC

m

2 

34 Electric field intensity due to an infinite sheet of charge is given by E=Є 0 / σ E=σ Є 0 E=σ/ Є 0 E=σ/2 Є 0



35 Electric field intensity between oppositely charged parallel plates E=Є 0 / σ E=σ Є 0 E=σ/2 Є 0 E=σ/ Є 0



A potential difference b/w two points is one volt. The amount of work

One coulomb one electron volt One erg

One joule



done in moving a charge of one coulomb from one point to another is

37 The SI unit of potential difference is Ampere Coulomb Joule volt



38 Electric potential energy per unit charge is also called Electric field Electric intensity Electric field

Electric

potential



39 The electric potential at a point due to a point charge is given by V= Kqr

2

Kq/r

2

Kqr Kq/r



Prof. Muhammad Shahid

Mirza (0321-7814911)

111-C Peoples Colony No.01 Refuljent Public School.

Faisalabad

Prof. Nveed Ahmad Janjua (0302-4425094)

Contacts: 0300-4590930, 0300-7618083 & 0331-

7814911

J S A c a d e m y F a i s a l a b a d Learn to Live, Live to Learn

JSA Series Important MCQS Physics Part-II

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40 Electric flux due to point charge is 1/ qЄ 0 Є 0 /q qЄ 0 /r q/ Є 0



41 If an electron is accelerated through a potential difference of one volt Ve

2

E/V V/2 Ve



it will acquire energy

42 Electron volt is the unit of Electric flux

Potential

Electric potential

energy



difference

Electron energy is one electron volt when it is accelerated through a

One erg One coulomb One joule

One volt



potential difference of

44 The magnitude of electric field between two point can be calculated by ∆V=Ed

2

∆V=Ed



∆V=d/E ∆V=E/d

the relation

45 The potential of all the points of a equi potential surface is infinite different zero same



46 Farad is the unit of Potential difference current charge capacitance



47 The SI unit of electric potential is Ampere Coulomb Joule volt



48 Unit of electric field intensity is Newton x meter meter/volt Volt x meter Volt/meter



.

49 A device used for storing charge is called

blogspot transistor com inductor resistor Capacitor

50 For a capacitor , the charge per unit volt is called Dielectric c nstant Charge density permittivity capacitance



51 Capacitance of a parallel plate capacitor depends on A Є 0 d All of above



52 Capacitance of a parallel plate capacitor is given by C=d/A Є 0 C=Є 0 /Ad^ C=A Є 0 d^ C=A Є 0 /d



TaleemTutor

.

Xerography



53 The copying process is called angiography topography photography

54 A 50μF capacitor has a potential difference of 8V cross it. The charge 6.25x

C 4x

C 4x

C (^) 4x

C

on the capacitor is

55 Capacitance of a parallel plate capacitor does not depend on A Є 0 d

Material of the

plates



Three capacitors of capacitance 1μ farad each are connected in series.

0.03 μF 9μF 3 μF

1/3 μF



Their equivalent capacitance is

57 Energy stored in a capacitor is given by the formula 2 CV

2

C/V

2

CV

2

CV

2

/2

If a slab of dielectric is inserted b/w the plates of a parallel plate

Becomes infinite

Remains

decreases

increases



capacitor connected across a battery. its stored energy constant

59 1 micro coulomb is equal to 10

coulomb 10

coulomb 10

coulomb 10

• coulomb



60 When a dielectric is placed in an electric field it Gets uncharged Gets unpolarized Gets charged Gets polarized



61 4 μF & 2 μF are connected in series, their equivalent capacitance is 2 μF 6 μ F 0.75 μF 1.3 μF



Two 50 μF capacitors are connected in parallel their equivalent

1 μF 25 μF 50 μF

100 μF



capacitance

63 In a charged capacitor energy resides in the form of Nuclear field Gravitational Magnetic flied Electric field



Prof. Muhammad Shahid Mirza (0321-7814911)

111-C Peoples Colony No.01 Refuljent Public School.

Faisalabad

Prof. Nveed Ahmad Janjua (0302-4425094)

Contacts: 0300-4590930, 0300-7618083 & 0331-

7814911

J S A c a d e m y F a i s a l a b a d Learn to Live, Live to Learn

JSA Series Important MCQS Physics Part-II

---

field

If a dielectric is inserted b/w the plates of a charged capacitor, its

Becomes infinite

Remains

decreases

increases



capacitance constant

Selenium is an insulator in the dark but when exposed to light it

Remains insulator semiconductor Super conductor

conductor



becomes

special dry black powder is spread over the drum of photocopier is

neutralizer Photo powder turner

toner



called

67 Photo copier and the inkjet printer are examples of electricity magnetism electromagnetism electrostatics



68 Since selenium becomes conductor in light it is called Photo diode Photo tube photocell

Photo

conductor



69 Charge on an electron was measured by Millikan in 1920 1909



1905 1900

70 Electric field intensity inside a hollow charged sphere is minimum infinity maximum zero



speed of charging and discharging of a capacitor depends on

charge

com

Potential

current

capacitance



resistance & difference

. 74 SI unit of capacitance is Volt/Coulombblogspot N/C volt Farad

72 In a charged capacitor the energy resides in Dielectric Positive plate Negative plate Field b/w plates



Electric flux due to a point charge q present inside a closed surface can

Lenz’s law Coulomb’s law Ohm’s law

Gauss’s law



be calculated by

.

TaleemTutor

The charge on the droplet in Millikan experiment is calculated by

Qε=V/mgd Qε=mg/dv Qε=d/mgv Qε=mgd/V

formula

76 The relation (∆V/∆r=V/d) represents Gauss’s law Electric flux Potential difference

Electric field

intensity



77 Farad = Joule/ coulomb Volt/Coulomb Coulomb x volt Coulomb/volt



78 Unit of capacitance is Joule/ coulomb Volt/Coulomb Coulomb x volt Coulomb/volt



79 Dielectric is also called Super conductor Semi conductor conductor insulator



Mechanical

electrical

80 If a charged body is moving against the electric field it will gain Potential energy Kinetic energy Potential

energy

energy



81 Xerography means average Breaking down Liquid writing Dry writing



82 The term RC has the same unit as that of (RC= ) 1/ t

2

t

2

1/t (^) t

83 One electron volt is equal to (^) 1.6x

J

 1.6x

19

J 6.25x

J 6.25x

18

J

84 Energy density in case of capacitor is always proportional to C E

2  V

2

Є 0

85 Presence of dielectric always

Increase the decrease the double the Does not affect

electrostatic force electrostatic electrostatic force the electrostatic

Prof. Muhammad Shahid Mirza (0321-7814911)

111-C Peoples Colony No.01 Refuljent Public School.

Faisalabad

Prof. Nveed Ahmad Janjua (0302-4425094)

Contacts: 0300-4590930, 0300-7618083 & 0331-

7814911

J S A c a d e m y F a i s a l a b a d Learn to Live, Live to Learn

JSA Series Important MCQS Physics Part-II

---

force



force

86 The electric field created by positive charge is Radially outward



Radially inward circular zero

87 The minimum charge on an object can not be less than 1C (^) 1.6x

C^ 1.6x

19

C none

Two point charges +2C and +6C repel each other if a charge 0f

0N



8x

9

N 9 12x

9

N

-2C is given to each of them then electrostatic force between them is (attractive)

108x10 N (repulsive) (attractive and

repulsive)

89 The unit of energy density of electric field is J/C J/V J/m

3  J/F

3

90 For the computation of electric flux, surface area should be Flat



Curved Inclined spherical

Ch#13(Current Electricity)

S.# QUESTIONS A B C D

1 Through metallic conductor the current is because of flow of photons neutrons Positive charges electrons



2 The charge per unit time through any cross-section of a conductor is

Potential

energycom (^) Electric power capacitance current



called

blogspot .

3 I= ∆Qε/∆I ∆t/∆Qε ∆Qεx∆t ∆Qε/∆t



4 One Coulomb/sec = Ohm capacitance volt ampere



5 S.I unit of electric current is Ohm coulomb voltage Ampere



TaleemTutor 

If 1 ampere current flows through 2m long conductor, the charge flow.

1 C 2 C 7200 C 3600 C



through this conductor in 1 hour will be

7 The graphical representation of Ohm’s law is hyperbola parabola Ellipse Straight line



8 ∆Qε= 1/(∆Qε/∆t) ∆I+∆t ∆t/∆I ∆Ix∆t



9 I= V

2

R VR R/V V/R



10 Ohm is the unit resistivity conductance current resistance



11 Ohm is defined as Coulomb / volt Volt / coulomb Volt x ampere Volt/ampere

12 V=IR represents Coulomb’s law Faraday’s law Ampere’s law Ohm’s law



13 If the resistance of a conductor is increased then current Becomes zero

Remains

increases

decreases



constant

14 R= LA/ρ ρ/LA A/ ρL ρL/A



15 ρ = R/AL LR/A L/RA AR/L



16 The resistance of a meter cube of a material is called its resistance conductance conductivity resistivity



17 Reciprocal of resistance is called capacitance resistivity conductivity conductance



18 SI unit of resistivity is 1/ Ohm-meter meter/ Ohm Ohm/meter Ohm-meter



A wire of uniform area of cross section “A”, length “L” and resistance “R”

Is one-fourth Becomes half doubles

Remains same



is cut into two equal parts. The resistivity of each part

Prof. Muhammad Shahid Mirza (0321-7814911)

111-C Peoples Colony No.01 Refuljent Public School.

Faisalabad

Prof. Nveed Ahmad Janjua (0302-4425094)

Contacts: 0300-4590930, 0300-7618083 & 0331-

7814911

J S A c a d e m y F a i s a l a b a d Learn to Live, Live to Learn

JSA Series Important MCQS Physics Part-II

---

20 The resistance of a conductor does not depend on its Area length temperature Mass



21 Reciprocal of resistivity of a material is called capacitance resistance conductance conductivity



22 The conductance of a conductor increases by

Increasing Decreasing area

Increasing length

Decreasing

temperature of cross-section temperature



Resistance of a substance of one meter in length and one square meter in

resistance conductance conductivity

resistivity



cross section is called

24 Which of the following materials is useful for making standard resistance Tungsten Copper Nichrome constantan



25 When the temperature of a conductor is increased its resistance Becomes Zero Remains same decreases increases



26 Resistance of a conductor increases with increase in

Area of cross-

diameter Mass

length



section

The resistance of the conductor increases due to the rise of temperature

Becomes zero

Remains

decreases

increases



of a conductor , because the collision cross section of the atoms unchanged

The current through a resistor of 100 Ohm when connected across a

0.45 A

com 200 A 220000 A

2.2 A



source of 220 V . blogspot

(Rt-R 0 )/R 0 t



29 The temperature coefficient of resistance α= (R -R 0 )/t^ (Rt-R 0 )/R 0 (Rt+R 0 )/R 0 t

30 The temperature coefficient of resistivity α= (ρ (^) t- ρ 0 )/ t (ρ (^) t- ρ 0 )/ ρ 0 (ρ (^) t+ρ 0 )/ ρ 0 t (ρ (^) t- ρ 0 )/ ρ 0 t



31 SI unit of temperature coefficient of resistivity is Ohm

Ohm K (^) K

• 

32 The potential difference across each resistance in series combination is

maximum zero same different



TaleemTutor

Two resistors of 2 ohm & 4 ohm are connected in parallel their equivalent

4 Ohm 6 Ohm 1.5 Ohm

1.33 Ohm



resistance is

Three resistors of resistance 2,3 and 6 Ohms are connected in parallel the

11 Ohm 3 Ohm 5 Ohm

1 Ohm



equivalent resistance will be

Three resistances 5000, 500 and 50 Ohms are connected in s ries across

10 mA 1 A 10 A

100 mA



555 volts main. The current flowing through them will be

36 Why should different resistances be added in series in a circuit

To decrease To increase

to divide voltage



None of these voltage voltage

37 P= I

2

/R RI

2

t I

2

V I

2

R



38 Heat generated by a 40 Watt bulb in one hour is 4800 J 1440 J 14400 J 144000 J



39 How will you calculate power from current I and Voltage V I

2

/R R/I

2

I

2

V VI



40 Electrical energy is measured in Kilo watt Horse power watt Kilowatt hour



A 100 watt bulb is operated by 200 volt, the current flowing through the

2.5 ampere Zero ampere 1 ampere

0.5 Ampere



bulb is

42 The resistance of a 60 watt bulb in a 120 volt line is 0.5 Ohms 2 Ohms 20 Ohms 240 Ohms



43 Electrical energy is given by the formula I

2

R VIt IRT I

2

Rt



Prof. Muhammad Shahid Mirza (0321-7814911)

111-C Peoples Colony No.01 Refuljent Public School.

Faisalabad

Prof. Nveed Ahmad Janjua (0302-4425094)

Contacts: 0300-4590930, 0300-7618083 & 0331-

7814911

J S A c a d e m y F a i s a l a b a d Learn to Live, Live to Learn

JSA Series Important MCQS Physics Part-II

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44 1 kilo watt hour is equal to 360000 J 3.6 x 10

5

J 3.6 x 10

7

J 3.6 x 10

6 J



45 If a 40 watt light bulb burns for 2 hours how much heat is generated 400 J 80 J 280 x 10

5

J 288 x 10

3 J



46 Which one of the following bulb has least resistance 100 watt 200 watt 500 watt 1000 watt



A fuse is placed in series with the line wire of house circuit to protect

Over heating High voltage high power

High current



against

A 1000 watt heater operates on a 220 volt line for one hour. The current

6.5 A 5 A 7 A

4.5 A



passing through the heater is

The electromotive force of a battery or cell is the voltage b/W its

Its internal Its internal

49 resistance is resistance is Circuit is closed Circuit is open



terminals when

maximum minimum

50 Electromotive force is given by the formula E=W

2

/q E=qW E=q/W E= W/ q



51 S.I unit of electromotive force is Ohm Coulomb farad volt



.

Magnetic flux Electric field Potential

52 Electromotive force is closely related to

com

Inductance

density intensity difference



53 By electromotive force S und is produced heat is produced Light is produced

Current is

produced



blogspot

TaleemTutor Internal resistance Internal

Battery is Battery is

54 Terminal potential difference of a battery is greater than its emf when

. of a battery is resistance of a

discharged charged



infinite battery is zero

Electrical energy Heat energy into

Electrical energy Chemical energy

55 Batteries convert into mechanical into electrical

into heat energy chemical energy

energy energy



56 The charge carriers in electrolyte are protons positive ions negative ions Both (b) and (c)^



57 Electronic current is due to flow of Positrons positive ions protons electrons



58 SI unit of conductance is K

-1 Ohm-meter Ohm mho



59 A conductor which strictly obeys ohm’s law is called

Electrolytic Supper

non-ohmic

Ohmic



resistor conductor

60 Semi-conductor diode is an example of

Electrolytic Supper

Ohmic device

Non-ohmic

resistor conductor device



61 The substances having negative temperature co-efficient are carbon germanium Silicon All of them



62 A carbon resistor consists of --------- colour bands 6 1 2 4



63 The tolerance of silver band is 5℅ ±20℅ ±10℅



±5℅

64 Rheostat can be used as a Current source Potential divider Variable resistor Both (b) and (c)



Prof. Muhammad Shahid Mirza (0321-7814911)

111-C Peoples Colony No.01 Refuljent Public School.

Faisalabad

Prof. Nveed Ahmad Janjua (0302-4425094)

Contacts: 0300-4590930, 0300-7618083 & 0331-

7814911