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Physics NEET MCQ
Quiz 4
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Q.1
A metallic shell has a point charge 'q' kept inside its cavity. Which one of the following diagram correctly represnets the electric lines of forces?
0%
a)
0%
b)
0%
c)
0%
d)
Explanation
electric field inside the metaliic portion is zero hence option (a) and (d) are incorrect.Electric field lines are normal to a surface. Hence option (b) is incorrect.Only option (c) represents the correct aswer Answer:(c)
Q.2
Six charges of equal magnitude, 3 positive and 3 negative are to be placed on PQRSTU corners of a regular hexagon, such that field at the centre is double that of what it would have been if only one +ve charge is placed at R [ IIT 2004]
0%
a) +, +, +, -, -, -
0%
b) -, +, +, +, -, -
0%
c) -, +, +, -, +, -
0%
d) +, -, +, -, +, -
Explanation
If opposite charges are kept on opposite verices electric field will be added . and if like charge are kept at vertices electric field will be cancelled. Thus option C is correct Answer: (c)
Q.3
A Gaussian surface in the figure is shown by dotted line. The electric field on the surface will be [ IIT 2004]
0%
a)due to q1 and q2 only
0%
b) due to q2 only
0%
c)zero
0%
d)due to all
Explanation
The flux through the Gaussian surface is due to the charges inside Gaussian surface. But electric field on the Gaussian surface will be due to the charges present on the Gaussian surface and outside it. It will be due to all the chargesAnswer: (d)
Q.4
Three infinitly long charge sheets are placed as shown in figure. the electric field at point P is [ IIT 2005]
0%
a)
0%
b)
0%
c)
0%
d)
Explanation
Drection of electric field produced at point P due to all the paltes in along negative x axis The total electric field E=E1 + E2 + E3 Answer: (c)
Q.5
A long, hollow conducting cylinder is kept coaxially inside another long, hollow conducting cylinder of larger radius. Both the cylinders are initially electrically neutral [ IIT 2007]
0%
a)A potential difference appears between the two cylinders when a charge density is given to the inner cylinder
0%
b) A potential difference appears between the two cylinders when a charge density is given to the outer cylinder
0%
c)No potential difference appears between the two cylinders when a uniform line charge is kept along the axis of the cylinder
0%
d)No potential difference appears between the two cylinders when same charge density is given to both the cylinders
Explanation
When a charge density is given to the inner cylinder, the potential developed at its surface is different from that on the outer cylinder. this is beacuse the potential decreases with distance from a charged conducting cylinder when the point of consideration is outside the cylinder. But when a charge densitty is given to the outer cylinder, it will chage its potential by the same amount as that of the inner cylinder. Therefore no potential difference will be produced between the cylinders in this case Answer:(a)
Q.6
Consider a neutral conducting spher. A positive point charge is placed outside the sphere. The net charge on the sphere is then [ IIT 2007]
0%
a) negative and distributed uniformly over the surface of the sphere
0%
b) negativ and appers only at the point on the sphere closer to the point charge
0%
c) negative and distributed non-uniformly over the entire surface of the sphere
0%
d) Zero
Explanation
When a positive point charge is placed outside a conducting sphere, a rearrangement of charges takes place on the surface. But the total charges on the sphere is zero as no charge has left or entered the sphere Answer: (d)
Q.7
A spherical portion has been removed from a solid surface having a charge distributed uniformly in its volume as shown in the figure. The electric field inside the empited space is .. [ IIT 2007]
0%
a)zero everywhere
0%
b) non-zero and uniform
0%
c)non-uniform
0%
d)zero only at its center
Explanation
Answer: (b)
Q.8
A proton is released from rest at a distance of 10-4Å from nucleus of mercury atom (Z=80). The kinetic energy of the proton when it is far away from the nucleus is
0%
a) 12eV
0%
b) 12 KeV
0%
c) 1.2MeV
0%
d) 12 MEV
Explanation
Kinetic energy of proton at inifinity=Potentil energy of proton at 10-4Å Answer: (d)
Q.9
positive and negative point charges of equal magnitude are kept at (0, 0, a/2) and ( 0, 0, -a/2) respectively. the work done by the electric field when another positive point charge is moved from ( -a, 0, 0) to ( 0, a, 0) is [ IIT 2007]
0%
a) positive
0%
b) negative
0%
c)zero
0%
d)depends on the path connectingthe initial and final positions
Explanation
Magnitude of position for the point is same thus potential at both the point is same , hence zero workAnswer: (c)
Q.10
Consider a system of three charges q/3, q/3 and 2q/3 placed at points A, B and C resppectively, as shown in the figure. Take O to be the centre of the circle of radius R and angle CAB=660°.. [ IIT 2008]
0%
a) The electric field at point O is q / 8πεoR2 directed along negative x-axis
0%
b) the potential energy of the system is zero
0%
c)the magnitude of the force between the charges at C and B is q2 / 54πεoR2
0%
d)the potential at point O is q/ 12πεoR
Explanation
option a : Electric field due to A and B at O is equal and opposite producing a resultant which is zero. The elctric field at O is due to charge at C=q / 6πεoR2 option is not correctOption b and d : Potential at O can not be zero as all the points are at equii distance from O and charges total is zon zero. Thus potential energy can not be zero. Option b and D are not correct option c: Magnitude of force between B and C is=Option is correct Answer:(c)
Q.11
A parallel plate capacitor C with plates of unit area and separation d is filled with liquid of dielectric constant K=The level of liquid is d/3 initially. Suppose the liquid level decreases at a constant speed v, the time constant as a function of time t is [ IIT 2008]
0%
a)
0%
b)
0%
c)
0%
d)
Explanation
Let the level of liquid at instant of time 't' be x. Then v=-dx/dt dx=-vdt Thus we can consider two cpacitor with distance (d-x) and x are connected in series . equivalent capacitance Time constant τ=R Ceq Answer: (a)
Q.12
Theree concentric metallic spherical shell of radii R, 2R, 3R are given charges Q1, Q2 and Q3 respectively. It is found that the surface charge densities of outer surfaces of the shells are equal. Then the ratio of the charges given to the shell Q1: Q2 : Q3 is [IIT 2009]
0%
a)1:2:3
0%
b) 1:3:5
0%
c)1:4:9
0%
d)1:8:18
Explanation
Due to induction and added charge :Sphere with radius 2R will have total charge Q1+ Q2 sphere of radius 3R is Q1 + Q2 + Q1Charge on sphere with radius R=Q1 since surface charge density of all the sphere is same Q1 : Q2:Q3=1:3:5Answer: (b)
Q.13
A disc of radius a/4 having a uniformly distributed charge 6C is placed in the x-y plane with its centre at ( -a/2, 0, 0). A rod of length a carrying a uniformly distributioed charge 8C is placed on the x-axis from x=a/4 to x=5a/Two point charges -7C and 3C are palaced at (a/4, -a/4, 0) and (-3a/4, 3a/4, 0), respectively. Consider a cubical surface formed by six surfaces x=± a/2, y=±a/2, x=± a/The electric flux throgh this cubical surface is [ IIT 2009]
0%
a) -2C/εo
0%
b) 2C/εo
0%
c) 10C/εo
0%
d)12C/εo
Explanation
From the figure it is cklaear that Half of the disc is eneclosed in cube thus charge eneclosed=3Csides of cubes are a/2Total lengeth of rod is a and charge is 8C and rod is palced at x=a/4 thus cube enclosed a/4 length=2C Charge 7C is at (a/4, -a/4, 0) which will be enclosed by cube Charge 3C is at (-3a/4, 3a/4, 0) is out of the cube Thus total charge enclosed=3C+2C-7C=-2C. Therefore the elctric flux through the cube is φ=-2C/εoAnswer: (a)
Q.14
Eight drops of mercury of equal radii and possessing equal charges combined to form a big drop. Then the capacitance of bigger drop compared to each individual drop is [ MNR 1987]
0%
a) 8 times
0%
b) 4 times
0%
c)2 times
0%
d)32 times
Explanation
Capacitance of spherical drop ∝ radiusvolume of big drop=8× volume of small drop Thus , radius of dig drop=2 × radius of small drop Capacitance of spherical drop ∝ radiusThus cpacitance of big drop=2× cpacitance of small drop Answer:(c)
Q.15
Two condensers of capacity 0.3µF and 0.6µF respectively are connected in series. The combination is connected across a potential of 6V. The ratio of energies stored by the condenser wil be [ MPPMT 1990]
0%
a) 1/2
0%
b) 2
0%
c) 1/4
0%
d) 4
Explanation
capcitors are connected in series thus charge on both capacitor is same Thus U1=q2 / 2C1 U2=q2 / 2C2 ∴ U1 / U2=C2 / C1 ∴ U1 / U2=2 Answer: (b)
Q.16
A parallel plate capacitor has a capacity C. The separation between plates is doubled and a dielectric medium is inserted between plates. The new capacitance is 3C. The dielectric constant of medium is
0%
a)1.5
0%
b) 3.0
0%
c)6.0
0%
d)12.0
Explanation
C'=KCAnswer: (c)
Q.17
If two conducting spheres are separately charged and then brought in contact
0%
a) the total energy of the two sphere is conserved
0%
b) the total charge on the spheres is conservered
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c)both the total energy and charge are conservered
0%
d)the final potential is always the mean of the original potential of the two spheres
Explanation
Answer: (b)
Q.18
A potential difference V is applied across two capacitors of capacitance C1 and C2 connected in series. Then the potential difference across C1 will be
0%
a) VC2 / C1
0%
c)VC2 / ( C1 + C2)
0%
d)VC1 / ( C1 + C2)
0%
b) V( C₁ + C₂) /C₁
Explanation
equivalante capacitance of combination=(C1C2) / ( C1 + C2) Thus charge on the combination q=(VC1C2) / ( C1 + C2) ∴ potential difference across C1=VC2 / ( C1 + C2) Answer:(c)
Q.19
Three capacitors of capacitance 3µF, 9µF and 18µF are connected in series and another time in parallel. The ratio of equivalent capacitance in the two case ( Cs / Cp) will be [ CPMT 1990]
0%
a) 1:15
0%
b) 15:1
0%
c) 1:1
0%
d) 1:3
Explanation
Cs=2µF Cp=30µFs / Cp)=1:15Answer: (a)
Q.20
The electric field between the two spheres of a charged spherical condenser: [ MPPMT 1994]
0%
a)is zero
0%
b) is constant
0%
c)increases with distance from the centre
0%
d)decreses with distance from the centre
Explanation
Answer: (d)
Q.21
The capacity of a parallel plate condenser is 5µF. When a glass plate is placed between the plates of the condenser, its potential difference reduces to 1/8 of the original value. The value of the dielectric constant of glass is [ MPPMT 1985]
0%
a) 1.6
0%
b) 8
0%
c)5
0%
d)40
Explanation
K=Vo / V=8Answer: (b)
Q.22
A glass slab is put within the plates of a charged parallel plate condenser. Which of the following quanities does not change? [ MPPMT 1998]
0%
a) energy of the condenser
0%
b) capacity
0%
c)intensity of electric field
0%
d)charge
Explanation
Answer:(d)
Q.23
An infinite number of identical capaciotrs each of capaciatance 1µF are connected as shown in figure. Then the equivalent capacitance between A and B is [ AP 1990]
0%
a)1 µF
0%
b) 2µF
0%
c)1/2µF
0%
d)∞
Explanation
Rows contains capaciors 1, 1/2, 1/4, 1/8, 1/16,... µFabove value capacitors are connected in parallel thus C=1+ (1/2)+(1/4)+(1/8)+(1/16) +...It is gemometric progression∴ C=first term/ (1 -difference) Here a=1 and r=1/2C=1/(1-1/2)=2µFAnswer: (b)
Q.24
A capacitor of capacitance 2µF is charged to a potential difference of 200 V. After disconnecting from battery, it is connected in parallel with an another uncharged capacitor. The common potential is 40V. The capacitance of the second capacitor is [ CPMT 1991]
0%
a) 2µF
0%
b) 4µF
0%
c)8µF
0%
d)16µF
Explanation
If V is common potential, From formula Answer: (c)
Q.25
Two point charges at a certain distance experience a force of 5N. Each charge is doubled in magnitude and distance between the two is halved. The interacting force would be
0%
a) 16 N
0%
b) 80 N
0%
c)5 N
0%
d)20 N
Explanation
Answer:(b)
Q.26
Four metallic plates each with a surface area of one side A, are palced at a distance d from each other. The plates are connected as shown in the figure. Then the capacitance of the system between a and b is :
0%
a)3εoA / d
0%
b) 2εoA / d
0%
c)2εoA / 3d
0%
d)3εoA / 2d
Explanation
Effective capacity between a and b is C/2+C=3C/2=3εoA/2d Answer:(d)
Q.27
Pulling the plates of charged capacitor apart
0%
a)increases the capacitance
0%
b) increases the potential difference
0%
c)does not affect potential difference
0%
d)decreases the potential difference
Explanation
Answer: (b)
Q.28
N identical spherical drop charged to same potential V are combined to form the big drop will be [ JIPMER 1998]
0%
a) V
0%
b) V×N
0%
c)V/N
0%
d)V ×N2/3
Explanation
Radius of big sphere R=N1/3 r Thus potential of big sphere=kNq/ ( N1/3r Potential=N2/3VAnswer: (d)
Q.29
The plates of a parallel plate capacitor are charged up to 100V. A 2mm thick plate is inserted between the plates, then to maintain the same potential difference, the distance between the capacitor plates is increased by 1.6 mm. The dielectric constant of the plates is [ MPPMT 1991]
0%
a) 5
0%
b) 1.25
0%
c)4
0%
d)2.5
Explanation
V=q/Clet charge on the capacitor after insertion of plate be q1 and capacitance be C1∴ q/C=q1 / C1let thickness of plate be t and dielectric constant be k then effective increase in the distance t/k let d' be the new distance between the plate then resultant effective distance=d'- t+ t/k Answer:(a)
Q.30
Four equal capacitors, each with a capacitance C are connected to a battery of e.m.f 10V as shown in figure. The midpont of the capacitor system is connected to earth. Then the potentials of B and D are respectively
0%
a) +10V, zero voly
0%
b) +5V , -5V
0%
c) -5V, +5V
0%
d) zero volt, 10 volt
Explanation
Total capacitance=C/4 Charge flowing through capacitors=CV=(C/4) × 10=10C/4=2.5C Potential at F is zero Potential across each capacitor between B and F is=v=q/C V=2.5C / C=2.5 V. Thus potential across two capacitor=2.5+2.5=+5V Similarly Potential across each capacitor Between F and D=-2.5 Thus potential across two capacitor=-2.5-2.5=+5V Answer: (b)
Q.31
Force of attraction between the plates of a parallel plate capacitor is
0%
a)
0%
b)
0%
c)
0%
d)
Explanation
let q be the charge on the plates We may consider secend plate is in the electric field of first. Thus second plate experience forec F=EqElectric field due to first plate E=σ/2εoK but σ=q/A Thus E=q/2εoAK F=q2/2εoAKAnswer: (a)
Q.32
six equal capacitors each of capacitance C are connected as shown in figure. Then equivalent capacitance between A and B is
0%
a) 6C
0%
b) C
0%
c)2C
0%
d)C/2
Explanation
Capacitor 1 to 5 forms a balanced Whetastone bridge. Current will not flow through capactor 3. Capacitor 1 and 4 are in series=C/2Capacitor 2 and 5 are in series=C/2Above combination is parallel to each other thus capacitance=C This resultant is parallel to capacitor 6 thus final resulatnet capacitance=C + C=2CAnswer: (c)
Q.33
A capacitor is charged by using a battery, which is then disconnected. A dielectric slab is then slipped between the plates which results in [ MPPMT 1995]
0%
a) reduction of charge on the plates and increase of potential difference across the plates
0%
b) increase in the potential difference across the plates, reduction in stored energy, but no change in the charge on the plates
0%
c)decrease in the potential difference across the plates, reduction in the stored energy, but no change in the charge on the plates
0%
d)none of the above
Explanation
Answer:(c)
Q.34
the combined capacit of the parallel combination of two capacitor is four times their combined capacity when connected in series. This means that [ EAMCET 1994]
0%
a)their capacites are equal
0%
b) their capacities are 1µF and 2µF
0%
c)their capacities are 0.5µF and 2µF
0%
d)their capacities are infinite
Explanation
Answer: (a)
Q.35
Figure a shows two capacitors connected in series and joined to a battery. The graph in figure (b) shows the variation in potential as one moves from left to right on the branch containing the capacitors if. [ MPPMT 1999]
0%
a) C1 > C2
0%
c)C1 < C2
0%
d)the information is not sufficient to decide the relation between C1 and C2
0%
b) C₁=C₂
Explanation
Answer: (c)
Q.36
The capacity of a parallel plate condenser depends on [ MPPMT 1994]
0%
a) the type of metal used
0%
b) the thickness of plates
0%
c)the potential applied across the plates
0%
d)the separation between the plates
Explanation
Answer:(d)
Q.37
The force between the plates of parallel plate capacitor of capacitance C and distance of separation of plate d with potential difference V between the plates, is [ MPPMT 1999]
0%
a) CV2 / 2d
0%
b) C2V2 / 2d2
0%
d)V2d / C
0%
c) C2V2 / d²
Explanation
The force between the plates of a parallel plate capacitor is given by Answer: (a)
Q.38
The numerical value of the charge on either plate of the capacitor C shown in the figure is
0%
a)CE
0%
b) CER1 / (R2 + r )
0%
c)CER2 / (R2 + r )
0%
d) CER1 / (R1 + r )
Explanation
there will be no current in the branch containing capacitor , as the resistance offered by the capacitor in D.C. circuit is infinite. Therefore current drawn from the battery.I=E / (R2 + r )Potential difference across the terminal of the battery V=E -Ir=E - [E / (R2 + r )]V=ER2 / (R2 + r )Charge on capacitor q=CV q=C×ER2 / (R2 + r )Answer: (c)
Q.39
Separation between the plates of a parallel plate capacitor is d and the area of each plate is A. When a slab of material of dielectric constant K and thickness t is introduced between the plates, its capacitance becomes [ MPPMT 1989]
0%
a)
0%
b)
0%
c)
0%
d)
Explanation
Electric field in air between the plates=E1=q / εoA Electric field in dielectric slab E2=q / KεoA The elctric field E1 between the plate exists in the distance (d-t) and E2 in the distance tHence , if the potential difference between the plates be V, thenV=E1(d - t) + E2t Answer: (c)
Q.40
Two metal spheres of capacitance C1 and C2 carry some charges. They are put in contact and then separated. the final charges Q1 and Q2 on then will satisfy [ MPPMT 1999]
0%
a)
0%
b)
0%
c)
0%
d)
Explanation
On contact both the spheres will have same potential V=Q1 / C1=Q2 / C2Thus Q1 / Q2=C1 / C2 Answer:(b)
Q.41
The capacity of paralel plate ccondenser is 10µF without dielectric. Dielectric of constant 2 is used to fill half the thickness between the plates the capacitance in µF is .. [ EAMCET 1995]
0%
a)10
0%
b) 20
0%
c)15
0%
d)13.33
Explanation
From the formulaAnswer: (d)
Q.42
An electric dipole of moment p is placed normal to the line of force of electric field E, then work done in defleting it through an angle of 180 degree is
0%
a) pE
0%
b) +2pE
0%
c) -2PE
0%
d) zero
Explanation
Answer: (d)
Q.43
SI unit of electric permitivity is
0%
a) N m2C-2
0%
b) Am-1
0%
c) N/C
0%
d) C2/ N m²
Explanation
Answer: (d)
Q.44
A finite ladder is constructed by connecting several sections of 2µF, 4µF capacitor combinations as shown in figure. It is terminated by a capacitor C. What value should be chosen for C, such that the equivalent capacitance of the ladder between the points A and B becomes independent of the number of sections in between? [ MPPMT 1999]
0%
a) 4 µF
0%
b) 2 µF
0%
c)18µF
0%
d)6 µF
Explanation
Effective capacitance between A and B is C . for the infinite long ladder. Thus by adding one extra unit will not affect the capacitance as shown in figureCapacitance Across A' and B' will be C Thus C across A and B is is in series with 4µF capacitot effective capacitance C'=4C/ 4+C C' is in parallel with 2µF and resultant of this combionation is C Answer: (a)
Q.45
The value of one Farad in e.s.u. will be [ PET 1998]
0%
a) 3×1010
0%
b)9×1010
0%
c)(1/9)×10-11
0%
d)(1/3)×1010
Explanation
Answer:(b)
Q.46
Capacitance of a capacitor made by a thin metal foil is 2µF. If the foil is folded with paper of thickness 0.15 mm, and dielectric constant of paper is 2.5, width of paper is 40 mm, then length o foil will be [ Raj. PET 1997]
0%
a) 0.34 m
0%
b) 1.33 m
0%
c) 13.4 m
0%
d) 339 m
Explanation
Answer: (d)
Q.47
Between the plates of parallel plate condenser there is 1mm thick paper of delectric constant . It it is charged at 100 V. the eletric field in Volt/metre between the plates of the capacitor is... [ MPPMT 1994]
0%
a) 100
0%
b) 100000
0%
c)25000
0%
d)400000
Explanation
E=V/d=100 / ( 10-3 )=100,000Answer: (b)
Q.48
Find the resultant capacitance between A and B [ Raj. PET 1997]
0%
a)(2/3)µF
0%
b) (8/3)µF
0%
c)(6/5)µF
0%
d)(7/3)µF
Explanation
Answer:(b)
Q.49
The area of plates of parallel plate condenser is A and the distance between the plate is 10 mm. There are two dielectric sheets in it, one of dielectricconstant 10 and thick ness 6 mm and other of dielectric constant 5 and thickness 4mm. The capacity of the condenser is [ MPPMT 1997]
0%
a) (12/35) εoA
0%
b) (2/3) εoA
0%
c) (5000/7) εoA
0%
d) (12/35) εoA
Explanation
Arrangement is equivalent to two capacitors in series having capacity Answer: (c)
Q.50
A condenser of capacity C1 is charged to a potential Vo. The eletrcostatic energy stored in it is Uo. It is connected to another uncharged condenser of capacity C2 in parallel. The energy dissipated in the process is [ MPPMT 1994]
0%
c)
0%
d)
0%
a)C2Uo) / (C₁ + C₂)
0%
b) C1Uo) / (C₁ + C₂)
Explanation
Loss of energyAnswer: (a)
0 h : 0 m : 1 s
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