MCQGeeks
0 : 0 : 1
CBSE
JEE
NTSE
NEET
English
UK Quiz
Quiz
Driving Test
Practice
Diagrams
Games
NEET
Physics NEET MCQ
Quiz 7
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
Q.1
A block of mass carrying a positive charge q is placed on a smooth horizontal table, which ends in a vertical wall situated at a distance d from block. An electric field E is switched on towards right. Assuming elastic collisions, find the time period of resultant oscillation
0%
a)
0%
b)
0%
c)
0%
d)
Explanation
two times the time taken by the block to collide on wall is the periodic time of oscillationAcceleration of the block a=qE/m d=(1/2)at2d=(1/2) (qE/m) t2 Thus 2t=√(8md/qE) Answer: (b)
Q.2
An electron having a charge -e located at A in the presence of a point charge +Q located at B is moved to a point C so that ABC is an equilateral . The work done in this process is
0%
a) Q / 4πεoAC
0%
b) Qe / 4πεoAC
0%
c)-Qe / 4πεoAC
0%
d)zero
Explanation
Electron is moving from A to C. A and C are equidistant from B and hence the potential at A due to charge +Q=potential at C due to the same charge. Since there is no potential difference between A and C no work is done in moving charge fro A to CAnswer: (d)
Q.3
An infinite number of charges each equal to q are placed along x-axis at x=1, x=2, x=4, x=8 and so on . the resultant potential at x=0 will be
0%
a) q / 2πεo
0%
b) q/4πεo
0%
c)q/8πεo
0%
d)q/εo
Explanation
Answer:(a)
Q.4
A solid conducting sphere of charge Q is surrounded by an uncharged concentric conducting spherical shell. The potential difference between the sphere and the shell is V. If the shell is now given a charge of -3Q, the new potential difference between them will be
0%
a) V
0%
b) 2V
0%
c) 4V
0%
d) -2V
Explanation
The potential difference depends only on the charge on the inner sphere Answer: (a)
Q.5
The effective capacitance between A and B is ( each capacitor is of 1µF)
0%
a)(15/2) µF
0%
b) (17/3)µF
0%
c)(13/8)µF
0%
d)(19/8)µF
Explanation
Circuit can be redrawn as Answer: (c)
Q.6
Two charged metal spheres of capacitance C1 and C2 are brought in contact and then separated. The final charges Q1 and Q2 on them will satisfy the condition
0%
a) Q1C1=Q2C2
0%
b) Q1C2 > Q2C1
0%
c)Q1C2 < Q2C2
0%
d)Q1C2=Q2C1
Explanation
both the spheres will save same potentials Q1 / C1=Q2 / C2 Answer: (d)
Q.7
How should 5 capacitors each of value 1µF be connected so as to produce a total capacitance (3/7)µF?
0%
a) Two capacitors in parallel and the combination in series with other three capacitors
0%
b) Three capacitors in a parallel and the combination in series with other two capacitors
0%
c) Four capacitors in parallel and the combination in series with fifth capacitor
0%
d) All capacitor in parallel
Explanation
Answer: (b)
Q.8
A neutron, a proton and an electron are placed in uniform electric field
0%
a)The forces acting on them will be equal
0%
b) Their accelerations may be equal
0%
c)Magnitude of acceleration may be equal
0%
d)Magnitude of acceleration will be different
Explanation
Charge on Neutron=0, Force=0mass of proton and electron is different hence acceleration differentAnswer: (d)
Q.9
A particle of mass m and charge q is fastened to one end of a string of length l. The other end of the spring is fixed to the point O. The whole system lies on a frictionless horizontal plane. Initially, the mass is at rest at A. A uniform electric field in the direction shown is then switched on. Then
0%
a) the speed of the particle when it reaches B is
0%
b) the speed of the particle when it reaches B is
0%
c)the tension in the string when the particle reaches B is qE
0%
d)the tension in the string when the particle reaches B is zero
Explanation
Work done=qE(l -lcosθ) Using work energy theorem Work done=ΔK.E.qE(l-lcosθ)=(1/2) mv2v=√(qEl/m) Answer: (b)
Q.10
The force between two point charges in vacuum is 15N, if brass plate is introduced between the two charges, then force between them will
0%
a) Become zero
0%
b) Remains same
0%
c)Becomes 30N
0%
d)Becomes 60N
Explanation
Answer: (a)
Q.11
A spark passes in air when the potential gradient at the surface of a charged conductor is 3×106 V/m. What must be the radius of an isolated metal sphere which can be charged to a potential of 3×106 volt before sparking into surrounding air ?
0%
a) 1 m
0%
b) 0.5 m
0%
c)8.85 m
0%
d)0.89 m
Explanation
We know that for charged sphere R=V/E on substituting values we get R=1mAnswer: (a)
Q.12
A 50µF capacitor is charged from 200V supply. After disconnected from supply, it is immediately connected in parallel with 30µF capacitor. The potential difference across the combination is
0%
a) 125 V
0%
b) 150 V
0%
c)175 V
0%
d)200 V
Explanation
use formula here V2=0 Answer:(a)
Q.13
An electric line of force is in the y-z plane is given by the equation y2 + z2=A particle with unit positive charge, initially at rest at the point y=0, Z=2 in the y-z plane
0%
a) will not move at all
0%
b) will move along z=0
0%
c) will move along a circular line of force
0%
d) will move along the line y=2
Explanation
line of force is circular with radius 2 . and charge is on the circle. charge will follow circular path of line of force Answer: (c)
Q.14
The capacitance of capacitor does not depend on
0%
a)the shape of the plates
0%
b) the size of the plates
0%
c)the charges on the plates
0%
d)the separation between the plates
Explanation
capacitance depends on the geometry of capacitorAnswer: (c)
Q.15
A and B are two thin concentric hollow conductors having radii a and b and charge q and Q respectively. Given that a > b and P is a point between the two spheres and distance of P from the common centre is ( b
0%
a) q+Q/r
0%
b) q/a + Q/r
0%
c) q/a + Q/b
0%
d)q/b + Q/a
Explanation
Point p is inner point for q thus potential at p=kq/a and outer point for Q potential=Q/r Total potential is algebraic sum Answer:(b)
Q.16
A parallel plate capacitor hs plates each of area A at separation d. It is charged to a potential V and then disconnected from the battery. If a slab of dielectric constant K is now introduced to fill the capacitor, the work done in the process is
0%
a)
0%
b)
0%
c)
0%
d)
Explanation
Change in energy=work Answer: (b)
Q.17
An air filed parallel plate capacitor is to be constructed which can store 12µC of charge when operated at 1200V. The dielectric strength of air is 3×106 V/m. What is the minimum area of the plates of the capacitor?
0%
a) 1 m2
0%
b) 0.45 m2
0%
c)1.5 m2
0%
d)1.2 m2
Explanation
Dielectric stretch of air E Answer: (b)
Q.18
A unit charge is given a displacement r=αi + βj in an electric field which has components E1 and E2 along x and y axes respectively. then work done is
0%
a)α(E₁ + E₂)
0%
b) β(E₁ + E₂)
0%
c)αE₂ + βE₁
0%
d)α E₁ + βE₂
Explanation
W=E . r Answer:(d)
Q.19
A charged sphere A having a radius 2cm is brought into constant with an uncharged sphere B having a radius 3cm. After the sphere are disconnected, the energy of B is 0.4J. The charge on the sphere A before contact is
0%
a)5.4 ×10⁻⁶C
0%
b) 2.7 ×10⁻⁶C
0%
c)1.8 ×10⁻⁶C
0%
d)3.6 ×10⁻⁶C
Explanation
Potential of sphere B=U=0.4J from the formula for energyAnswer: (b)
Q.20
The distance between the plates of a charged parallel plate capacitor disconnected from the battery is d=5 cm and the intensity of the electric field E=300V/m. An uncharged metal bar 1cm thick is introduced into the capacitor parallel to its plates. The ratio of the potential difference between the plates of the capacitor before and after the bar is introduced is
0%
a) 1.8
0%
b) 1.65
0%
c)1.25
0%
d)1.5
Explanation
Initial potential difference Vi=300 × 5=1500 V Since charge is same after introduction of metal plate Ci Vi=Cf Vf Answer: (c)
Q.21
Two uniformly large plane sheets S1 and S2 having charge densities σ1 and σ2 ( σ1 > σ2) are placed at a distance d parallel to each other. A charge qo is moving along a line of length a (a < d) at an angle of 45° with the normal to SThe work done by the electric field is
0%
a)
0%
b)
0%
c)
0%
d)
Explanation
Electric field between parallel plate=( σ1 - σ1 ) / 2εo Work=Eq× displacement W=Eq × acos45 Answer:(c)
Q.22
A particle of mass m and charge q starts moving from rest along a straight line in an electric field E=Eo - αx where α is positive constant and x is the distance from starting point. Find the distance traveled by the particle till the moment it comes to instantaneous rest
0%
a) 2Eo / α
0%
b) Eo / α
0%
c) Eoq/ m
0%
d) Eo / Q
Explanation
Force F=qE Thus a=qE/m but electric field is position dependent thus we have to integrate the function Answer: (a)
Q.23
Point charges q, -q, 2Q and Q are placed in order at the corners A, B, C and D of a square of side 2b. If the field at the midpoint CD is zero, then q/Q is
0%
a)1
0%
b) 2
0%
c)2√2 / 5
0%
d)5√5 /2
Explanation
Since the field at the midpoint CD is zero, then along CD EBcosθ + ED + EAcosθ - EC=0Answer: (d)
Q.24
An infinite plane sheet of positive charge has surface charge density σ A metallic ball of mass m and charge +Q is attached to a thread and is tied to a point A on the sheet PQ. The angle θ made by the string with the plane sheet is
0%
a)zero
0%
b)
0%
c)
0%
d)
Explanation
Electric field due to infinite plane sheet of positive charge E=σ/ 2ε0from figure Tcosθ=mg and Tsin θ=Qσ/ 2ε0∴ tanθ=Qσ/ (2ε0mg)Answer: (a)
Q.25
Four metallic plates, each with surface area of oneside A, are palced at distance d from each other. The plates are connected as shown in the adjoining figure. Then the capacitance of the system between a and b is
0%
a)
0%
b)
0%
c)
0%
d)
Explanation
To solve this type of questions number the platethen arrange them. Arrangement has three capacitors between 1 - 2, between 2 - 3 m, between 3 - 4 Effective capacity between A and B is C/2 +C=(3/2) C thus option " d " is correctAnswer: (d)
Q.26
A parallel plate capacitor of capacitance C is connected to a battery and is charged to a potential difference V. Another capacitor of capacitance 2C is similarly charged to potential difference 2V. The charging battery is now disconnected and the capacitor are connected in parallel to each other in such a way that the positive terminal of one is connected to the negative terminal of the other. The final energy of the configuration is
0%
a)zero
0%
b) (3/2)CV2
0%
c)(25/6)CV2
0%
d)(9/2)CV2
Explanation
Charge on first capacitor Q1=CV Charge on second capacitor Q2=4CVBy the law of conservation of charge total charge of the system=CV+ 4V=5CV Total capacitance of system=C+2C=3C If V' is potential difference across each capacitor Then V'=CV/3C=(5/3)VNow energy of the configuration=(1/2)CV'2 + (1/2) 2C V'2=(3/2)CV'2=(3/2)C(5V/3)2=25CV2 / 6 Answer:(c)
Q.27
Consider the situation shown in figure. The capacitor A has a charge q on it whereas B is uncharged. The charge appearing on the capacitor B a long time time after the switch is closed is
0%
a) zero
0%
b) q/2
0%
c) q
0%
d) 2q
Explanation
since circuit is not closed , negative charge is bound, no charge can flow to the condenser B. Hence charge on B is zero Answer: (a)
Q.28
In figure shown the electric lies of force emerging from charged body. IF the electric fields at A and B are EA and EB respectively. IF the distance between A and B is r then
0%
a)EA > EB
0%
b) EA < EB
0%
c)EA=EB
0%
d)EA=(EB) / r²
Explanation
Answer: (a)
Q.29
The enrgy U stored in the electric field produced by a metal sphere of radius R containing a charge Q is
0%
a) Q / 8εoR2
0%
b) Q2 / 8πεoR
0%
c)Q / 8πεoR
0%
d)Q2 / 8πεoR2
Explanation
E=kQ/r2 Energy density ρ(r)=(1/2)εoE2 Energy stored in volume dv=dU=ρ(r)dVTotal energy stored is U=Answer: (b)
Q.30
A charge 10-9C is located at the origin in free space and another charge Q at (2, 0, 0). If the X-componant of the electric field at (3, 1,1) is zero, calculate the value of Q
0%
a) -0.42 nC
0%
b) 0.72nC
0%
c) -0.92nC
0%
d) -0.72C
Explanation
From diagram As vector addition of X-componant of electric fields=0 Answer: (a)
Q.31
In a parallel plate capacitor, the plate separation of 10mm is very small compared with the size of the plates. A potential difference of 5.0kV is maintained across the plates. The electric field intensity between the plates is
0%
a)500 V/m
0%
b) 2.5×105 V/m
0%
c)5×105 V/m
0%
d)2.5×1013 V/m
Explanation
Use E=V/d Answer: (c)
Q.32
A point charge q is located at the centre O of a spherical uncharged conducting layer provided with a small orifice as shown in figure. The inside and outside radii of the layer are equal to a and b respectively. What amount of work has to be performed to transfer the charge q slowly from the point O to infinity through the orifice
0%
a)
0%
b)
0%
c)
0%
d)
Explanation
As positive charge is kept in cavity negative charge will induce on the inner surface and positive charge will induce on outer surface Thus Initial potential energy=Potential at O due to -q on inner surface + potential at o due to +q on outer surface Initial potential=V And final potential at infinity=0 Average potential energy=(Initial potential + final potential / 2=V/2 Work done=-Average potential energy ×q [ Negative sign as work done is against electric field] Answer: (a)
Q.33
A particle of mass m with charge q > 0 and initial kinetic energy K is projected from infinity towards a heavy nucleus of charge Q assumed to have a fixed position. If the aim is perfect, how close to the centre of the nucleus is the particle when it comes instantaneously to rest?
0%
a)
0%
b)
0%
c)
0%
d)
Explanation
Using law of conservation of energy ΔP.E=ΔK.E Answer:(a)
Q.34
The flow of charge through switch S when it is closed is
0%
a) zero
0%
b) q/4
0%
c) 2q/3
0%
d) q/3
Explanation
When switch is closed, the circuit reduces to both capacitor have same capacitance with same charge and same polarity ∴ no charge flows through the switch Answer: (a)
Q.35
The equivalent capacitance between point A and B shown in figure will be...( each capacitor is of capacitance 3µF)
0%
a)9 µF
0%
b) 1µF
0%
c)4.5µF
0%
d)6µF
Explanation
circuit is for parallel combination equivalent capacitance=9µFAnswer: (a)
Q.36
Point charge q moves from point P to point S along the path PQRS in uniform electric field E parallel to the positive direction of the x-axis. The coordinates of point P, Q, R and S are (a, b, 0) (2a, 0, 0), (a, -b, 0) and (0, 0, 0) respectively. The work done by the field in the above process is given by the expression
0%
a) qaE
0%
b) -qaE
0%
d)3qE(√[a2b2)]
0%
c)q(√(a²+b²)E
Explanation
Electrostatic force is conservative in natureThe net work done is work done in moving the charge from S to P W=F.SPW=(qEi)(ai + bj)W=qEa Thus work from PtoS=-qEaAnswer: (b)
Q.37
Which of the following statement regarding electrostatics is wrong
0%
a) Charge is quantized
0%
b) charge is conserved
0%
c) There is no field near an isolated charge at rest
0%
d) A stationary charge produces both electric and magnetic fields
Explanation
Answer: (d)
Q.38
The dielectric constant of water is
0%
a)1
0%
b) 40
0%
c)81
0%
d)0.3
Explanation
Answer: (c)
Q.39
A very long uniform charged thread oriented along the axis of a circle of radius R rests on its centre with one of the ends. The charge of the thread per unit length is equal to λ. Find the flux of electric field across the circular area due to charges on the thread
0%
a) R/2ε0
0%
b) λR/ε0
0%
c)λR/2ε0
0%
d)λR/ε02
Explanation
Electric field due to the semi infinite wire at elementary circle of radius r Answer:(c)
Q.40
The potential V(x, y) of an electrostatic field E=2axyi +a(x2 -y2)j. Where a is constant
0%
a) Vo +(a/3)y3 - ayx2
0%
b) Vo -(a/3)y3 + ayx2
0%
c) Vo +(a/3)y3 + ayx2
0%
d) Vo -(a/3)y3 - ayx2
Explanation
As -dV=E.dr=dxi + dyj -dv=[2axyi +a(x2 -y2)j].(dxi+dyj)-dV=2ayx dx + a(x2 -y2)dy Answer: (a)
Q.41
A charge q is placed at O in the cavity in a spherical uncharged conductor. Point S is outside the conductor. If the charge is displaced from O towards S still remaining with in the cavity
0%
a)electric field at S will increase
0%
b) electric field at S will decrease
0%
c)electric field at S will first increase and then decrease
0%
d)electric field at S will not change
Explanation
Charge on sphere B is uniformly distributed whatever be the position of q ∴ Es will not changeAnswer: (d)
Q.42
For the circuit shown, which of the following statement is true?
0%
a) With S1 closed, V1=15V, V2=20 V
0%
b) With S3 closed, V1=V2=25V
0%
c)With S1 and S2 closed V1=V2=0
0%
d)With S1 and S3 closedV1=30V, V2=20V
Explanation
When S1 and S3 are close, the circuit is still open. So V1=30V and V2=20VAnswer: (d)
Q.43
In the given circuit, with steady current, the potential drop across the capacitor C must be
0%
a) V
0%
b) V/2
0%
c)V/3
0%
d)2V/3
Explanation
In steady state no current flows through the capacitor. Hence resistance in loop=3R current through the loop=V/3RNow potential difference across V and R=Potential difference across V and C∴ V + (V/3R)R=V + VC VC=V/3 Answer:(c)
Q.44
A capacitor of capacitance C1=1µF withstands a maximum voltage V1=6kV while another of capacitance C2=2µF withstands a maximum voltage V2=4kV. If they are connected in series, maximum voltage the system can withstand is
0%
a) 15kV
0%
b) 12kV
0%
c) 9kV
0%
d) 6kV
Explanation
If V' is the maximum voltage that the system will withstand, when connected in series Form formula Answer: (c)
Q.45
A ball of radius R is uniformly charged with the volume density ρ. Find the flux of the electric field strength vector across the ball's section formed by the plane located at a distance ro < R from the centre of the ball
0%
a)
0%
b)
0%
c)
0%
d)
Explanation
Electric field inside a uniformly charged solid sphere at a point P distance r from Centrex E=ρx / 3εo Flux through the elementary ring of radius, ds=2πydy dφ=E.dscosθ and MN=√(R2 - rr)Answer: (a)
Q.46
Consider the arrangement of three plates X, Y, and Z each of area A with separation between successive plates being d. The energy stored when the plates are fully charged is
0%
a) εoAV2 / 2d
0%
b) εoAV2 / d
0%
c)2εoAV2 / d
0%
d)3εoAV2 / 2d
Explanation
Two capacitors are in parallel, energy stored U=(1/2)×2C×V2=CV2and C=Aεo / d Answer: (b)
Q.47
A parallel plate air capacitor is connected to battery. The charge, voltage, electric field and energy associated with this capacitor are given by Qo, Vo Eo and Uo respectively .A dielectric slab is now introduced to fill the capacitor with the battery connected. The corresponding quantities are now Q, V, E and U respectively. Then
0%
a)Q > Qo
0%
b) E > Eo
0%
c)Uo >U
0%
d)V > Vo
Explanation
We know that by introduction of dielectric and battery is connected then V=Vo C > Co U > Uo E < Eo and Q > Qo Answer:(a)
Q.48
The air plate can sustain an electric field of 3×106 V/m without breakdown of its insulation. The maximum amount of electrostatic energy per unit volume of air that can be stored in a parallel plate air capacitor is nearly
0%
a) 20 J/m³
0%
b) 40 J/m³
0%
c) 60 J/m³
0%
d) 80 J/m³
Explanation
Energy density=(1/2)εoE2 Answer: (b)
Q.49
A parallel plate capacitor is filled by a dielectric whose permitivity varies with applied voltage according to the equation εo=αV where α=1(Volt)-Another identical capacitor without dielectric charged to a voltage Vo=20 volt is connected in parallel to the first "non linear" uncharged capacitor. The final voltage across the capacitor is
0%
a)20 V
0%
b) 10 V
0%
c)5 V
0%
d)4 V
Explanation
Let Co is the capacitance of the capacitor without a dielectric. Change on the capacitor Q=CVo When capacitor of capacitance Co is connected to non-linear capacitor common potential be V then charge on Co q=CoV Capacitance of non linear capacitor C=εCo C=(αV)Co The charge on non-linear capacitance q'=((αV)Co) V=αCoV2 From conservation of charge Q=q + q' CoVo=CoV +αCoV2 V2 + V - Vo=0 V=- 5 volts or +4 volts Possible is V=4 voltsAnswer: (d)
Q.50
The ratio of electric force to gravitational force acting between two electrons will be
0%
a) 1×1036
0%
b) 2 × 1039
0%
c) 6 × 1045
0%
d)4 × 1042
Explanation
Answer:(d)
0 h : 0 m : 1 s
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
Report Question
×
What's an issue?
Question is wrong
Answer is wrong
Other Reason
Want to elaborate a bit more? (optional)