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Physics NEET MCQ
Quiz 3
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Q.1
A large circular ring has a uniform positive charge distribution. A negative charge on the axis and close to the centre is related from the rest. The negative charge
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a) has a uniform acceleration
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b) move towards the ring
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c)remains at rest
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d)executes a simple harmonic motion
Explanation
From the formula for for electric fieldF ∝ -x ∴ motion is S.H.MAnswer: (d)
Q.2
The speed of electron when accelerated through a potential difference of 5×105 volt is nearly
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a)4.4×108 m/s
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b) 2.59×108 m/s
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c)4.4×108 cm/s
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d)2.59×108 cm/s
Explanation
Relativistic K.E of electronK=(m - mo) C2Answer: (b)
Q.3
The figure shows some of the electric field lines corresponding to an electric field. The figure suggests .. [ MPPMT 1999]
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a) EA > EB > EC
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b) EA=EB > EC
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c)EA=EC > EB
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d)EA=EC < EB
Explanation
Distance between electric field lines at A nd C is less compared to point B. Thus At point A and point C electric field is equal but at point B electric field is lessAnswer: (c)
Q.4
An electric dipole is placed in an electric field generated by a point charge [ MPPMT 1999]
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a) the net electric force on the dipole must be zero
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b) the net electric force on the dipole may e zero
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c)the torque on the dipole due to the field must e zero
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d)the torque on the dipole due to the field may be zero
Explanation
Answer:(d)
Q.5
Two balls with equal charges are in vessel with ice at -10°C at a distance of 25cm from each other. On forming water at 0°C, the balls are brought nearer to 5cm for the interaction between them to be same. If the dielectric constant of water at 0°C is 80, the dielectric constant of ice at -10°C is
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a) 40
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b) 3.2
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c) 20
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d) 6.4
Explanation
In Ice In waterfrom above equations K(0.25)2=80(.05)2 K=3.2Answer: (b)
Q.6
A free proton and free α particle initially at a separation of 1Å are released, the kinetic energy of proton and that of &alpha-particle when at infinite distance, bear a ratio
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a)1:1
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b) 1:2
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c)1:4
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d)4:1
Explanation
From the law of conservation of momentumKinetic energy Kα=½ mαvα 2Kp=½ mpvp 2Answer: (d)
Q.7
The figure below shows two equipotential surface in X-Y plane for an electric field. The scales are marked. The x-component and y component of electric field in the region of the space where these equipotential lines exist, are respectively
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a) +100 Vm-1 , -200Vm-1
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b) -100 Vm-1 , +200Vm-1
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c)+200 Vm-1 , 100Vm-1
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d)-200 Vm-1 , -100Vm-1
Explanation
Answer: (b)
Q.8
On moving a charge of 20 C by 2cm, 2J of work is done, then the potential difference between the points is [ AIEEE 2002]
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a) 0.1V
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b) 8.0V
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c)2V
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d)0.5 V
Explanation
V=W/q V==2/20=0.1V Answer:(a)
Q.9
If there are n capacitors in parallel connected to V volt source., then the energy stored is equal to [ AIEEE 2002]
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a) CV
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b) ½ n CV2
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c) CV2
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d) (1/2n) CV2
Explanation
total capacitance=nC thus option 'b' is correct Answer: (b)
Q.10
A charged particle q is placed at the centre O of cube of length L ( ABCDEFGH). Another same charge q is placed at a distance L from O. Then the electric flux through ABCD is [ AIEEE 2002]
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a)q/ 4ΠεoL
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b) zero
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c)q/ 2ΠεoL
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d)q/ 3ΠεoL
Explanation
Both the charges are identical and placed symmetrically about ABCD. The flux passing through ABCD due to both the charges are equal but opposite in direction. Therefore resultant flux is zeroAnswer: (b)
Q.11
If the electric flux entering and leaving an enclosed surface respectively is φ1 and φ2, the electric charge inside the surface will be [ AIEEE 2003]
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a) ( φ2 - φ1)εo
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b)( φ2 + φ1) / εo
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c)( φ2 - φ1) / εo
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d)( φ2 + φ1)εo
Explanation
The flux entering an enclosed surface is taken as negative and the flux leaving the surface is taken as positive by convention. Therefore the net flux leaving the enclosed surface=φ2 - φ1∴ the charge enclosed in the surface by Gauss's law isq=εo (φ2 - φ1) Answer:(a)
Q.12
A sheet of aluminum foil of negligible thickness is introduced between the plates of a capacitor. The capacitance of the capacitor [ AIEEE 2003]
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a) decreases
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b) remains unchanged
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c) becomes infinite
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d) increases
Explanation
Answer: (b)
Q.13
A thin spherical conducting shell of radius R has a charge q. Another charge Q is placed at the centre of the shell. The electrostatic potential at a point P a distance R/2 from the centre of the shell is [ AIEEE 2003]
0%
a)
0%
b)
0%
c)
0%
d)
Explanation
Electrostatic potential due to charge Q placed at the centre of the spherical shell at point P is Electric potential due to charge q on the surface of the spherical shell at any point inside the shell is ∴ The net electric potential at point P is V=V1 + V2 Answer: (c)
Q.14
Three charges -q1, +q2 and -q3 are placed as shown in the figure. The x-component of the force on -q1 is proportional to ..[ AIEEE 2003]
0%
a)
0%
b)
0%
c)
0%
d)
Explanation
Force on charge q1 due to q2 is Force on the charge q1 due to q3 is The X-component of the force Fx on q1 is F12 +F13sinθ Answer:(b)
Q.15
Two spherical conductors B and C having equal radii and carrying equal charges on them repel each other with a force F when kept apart at some distance. A third spherical conductor having same radius as that B but uncharged is brought in contact with B, then brought in contact with C and finally removed away from both. The new force of repulsion between B and C is [ AIEEE 2004]
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a) F/8
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b) 3F/4
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c) F/4
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d) 3F/8
Explanation
Charge on B and C be q Sphere B is brought in contact with uncharged sphere thus on separation charge on C=q/2 Now C is brought in constant with third sphere thus charge on C=(q+q/2) / 2=3q/4 Answer: (d)
Q.16
A charged particle 'q' is shot towards another charged particle 'Q' which is fixed, with a speed 'v'. It approaches 'Q' up to a closest distance r and then returns. If q were given a speed of 2v the closest distance of approach would be [ AIEEE 2004]
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a)r/2
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b) 2r
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c)r
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d)r/4
Explanation
For closest approach Kinetic energy=potential energyAnswer: (d)
Q.17
Four charges equal to -Q are placed at the four corners of a square and a charge q is at its centre. If the system is in equilibrium the value of q is [ AIEEE 2004]
0%
a)
0%
b)
0%
c)
0%
d)
Explanation
To make system equilibrium resultant force on charge on any corner should be zero F1=F2=kQ2 / a2 F1 and F2 are perpendicular to each other other thus resultant F'=√2 ( kQ2 / a2)F3=kQ2 / 2a2F' and R 3 are in same direction thus F"=F'+F3 Now F"=F4Answer: (b)
Q.18
Two point charges +8q and -2q are located at x=0 and x=L respectively. the location of a point on the x axis at which the net electric field due to these two point charges is zero is [ AIEEE 2005]
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a) L/4
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b) 2L
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c) 4L
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d) 8L
Explanation
Answer: (b)
Q.19
Two think rings each having a radius R are placed at a distance d apart with their axes coinciding. The charges on the two rings are +Q and -Q. The potential difference between the centres of the two rings is [ AIEEE 2005]
0%
a)
0%
b)
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c)
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d)zero
Explanation
Potential at first ring V1=Vself + Vdue to 2 Potential at second ring V2=Vself + Vdue to 1 ΔV=V1 - V2 Answer: (a)
Q.20
A parallel plate capacitor is made by staking n equally spaced plates connected alternatively. If the capacitance between any two adjacent plates is C then the resultant capacitance is [ AIEEE 2005]
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a) ( n+1) C
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b) ( n-1)C
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c)nC
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d)C
Explanation
As n plates are joined , it means (n-1) combination joined in parallel∴ resultant capacitance=( n-1) CAnswer: (b)
Q.21
A charged ball B hangs from a silk thread S, which makes an angle θ with a large charged conducting sheet P, as shown in the figure. The surface charge density σ of the sheet is proportional to [ AIEEE 2005]
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a)cotθ
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b) cosθ
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c)tanθ
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d)sinθ
Explanation
From above figure Answer:(c)
Q.22
A fully charged capacitor has a capacitance 'C'. It is discharged through a small coil of resistance wire embedded in a thermally insulated block of specific heat capacity 's' and mass 'm'. If the temperature of the block is raised by ΔT, the potential difference 'V' across the capacitance is [ AIEEE 2005]
0%
a)
0%
b)
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c)
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d)
Explanation
Energy of capacitor=energy absorbed Answer: (c)
Q.23
An electric dipole is placed at an angle of 30° to a non-uniform electric field. The dipole will experience [ AIEEE 2006]
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a)a translational force only in the direction of the field
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b) a translational force only in a direction norml to the direction of the field
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c)a torque as well as translational force
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d)torque only
Explanation
Electric field is non uniform , two forces acting on the dipole are unequal and there line of action is not passing through a single point. There fore dipole will have linear force and torque Answer: (c)
Q.24
Two spherical conductors A and B of radii 1mm and 2mm are separated by a distance of 5cm and are uniformly charged. If the spheres are connected by the conducting wire then in equilibrium condition, the ratio of the magnitude of the electric field as the surfaces of the spheres A and B is [ AIEEE 2006]
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a) 4:1
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b) 1:2
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c)2:1
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d)1:4
Explanation
After connecting the sphere charge will flow from higher potential sphere to low potential sphere till potential of both the sphere become equalsince distance is very large as compared to their diameters, the induced effects may be ignoredLet Q1 and Q2 be the charges on the spheres then The ratio of electric fields Answer:(c)
Q.25
An electric charge 10-3 µC is placed at the origin ( 0, 0) of X-Y co-ordinate system. Two points A and B are situated at ( √2, √2) and ( 2, 0) respectively. The potential difference between the points A and B will be [ AIEEE 2007]
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a) 4.5 volts
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b) 9 volts
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c) zero
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d) 2 volt
Explanation
Potential at point due to point charge ∝ (1/r) Point charge is situated at (0,0) For point A (√2, &radic2) , r=2, VA ∝ 1/2 For point B ( 2,0) , r=2 VB ∝ 1/2 Thus potential at both the point will be same. Potential difference between point A and B=0 Answer: (c)
Q.26
Charges are placed on the vertices of a square as shown. Let E be the electric field and V be the potential at the centre. If the charges on A and B are interchanged with those on D and C respectively, then [ AIEEE 2007]
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a)E changes, V remains unchanged
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b) E remains unchanged, V changes
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c)both E and V changes
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d)E and V remain unchanged
Explanation
AS shown in figure, the resultant electric field after interchange will be same in magnitude but opposite in directions. Potential will be the same in both cases as it is a scalar quantityAnswer: (a)
Q.27
The potential at a point x ( measured in µm) due to some charges situated on the x-axis is given by V(x)=20 / (x2 - 4) voltThe electric field E at x=4µm is given by{AIEEE 2007]
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a) (10/9) V/µm and in the +ve x direction
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b) (5/3) V/µm and in the -ve x direction
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c)(5/3) V/µm and in the +ve x direction
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d)(10/9) V/µm and in the -ve x direction
Explanation
Positive sign indicates that E is in +ve directionAnswer: (a)
Q.28
A parallel plate condenser with a dielectric of dielectric constant K between the plates has a capacity C and is charged to a potential V volt. The dielectric slab is slowly removed from between plates and then reinserted. The net work done by the system in this process is [ AIEEE 2007]
0%
a) zero
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b)
0%
c)
0%
d)
Explanation
The potential energy of a charged capacitor is given by U=Q2 / 2CIf a dielectric slab is inserted between the plates, the energy is given by Q2 / 22KC, where K is dielectric constant Again, when the dielectric slab is removed slowly its energy increases to initial potential energy. Thus work done is zero Answer:(a)
Q.29
If gE and gM are the acceleration due to gravity on the surface of the earth and the moon respectively and if Millikan's oil drop experiment could be performed on the two surfaces, one will find the ratioelectronic charge on moon / electronic charge on earth=[ AIEEE 2007]
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a) gM / gE
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b) 1
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c) 0
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d) gE / gM
Explanation
Electronic charge does not depend on accleration due to gravity as it is a universal constant. So, electronic charge on earth=electronic charge on moon ∴ required ratio=1 Answer: (b)
Q.30
A parallel plate capacitor with air between the plates has capacitance of 9pF. The separation between its plates is 'd'. The space between the plates is now filled with two dielectrics. One of the dielectric has dielectric constant k1=3 . and thickness d/3 while the other one has dielectric constant k2=6 and thickness 2d/Capacitance of capacitor is now [ AIEEE 2008]
0%
a)1.8 pF
0%
b) 45 pF
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c)40.5 pF
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d)20.25 pF
Explanation
The given capacitance is equal to two capacitors connected in series Answer: (c)
Q.31
A charge Q is placed at each of the opposite corners of a square. A charge q is placed at each of the other two corners. If the net electrical force on Q is zero, then Q/q equals [ AIEEE 2009]
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a) -1
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b) 1
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c)- 1/√2
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d)-2√2
Explanation
Let F be the force between Q and Q. Therefore force between Q and q should be attractive to make resultant force on Q zero let force between Q and q be F' and resultant of Two F' forces be F"Thus F"=F for equilibriumAnswer: (d)
Q.32
A thin spherical shell of radius R has Q spread uniformly over its surface. Which of the following graphs most closely represents the electric field R(r) produced by the shell in the range 0≤ r < ∞, where r is the distance from the centre of shell? [ AIEEE 2008]
0%
a)
0%
b)
0%
c)
0%
d)
Explanation
Electric field inside the shell is zero, while out side E ∝r2Thse characteristics are represented by graph (a) Answer:(a)
Q.33
There exists a non-uniform electric field along x axis as shown in figure. the field increases at uniform rate along +Ve x-axis. A dipole is placed inside the field as shown. For the dipole which of the following statement is true
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a)Dipole moves positive x-axis and undergoes a clockwise rotation
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b) Dipole moves along negative x-axis and undergoes a clockwise rotation
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c)Dipole moves along positive x-axis and undergoes an anticlockwise rotation.
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d)Dipole moves along negative x-axis and undergoes an anticlockwise rotation
Explanation
Force on negative charge is in the direction of negative x axis and is more than the force on positive charge. dipole will move along negative x-axis with anticlockwise motionAnswer: (d)
Q.34
Two spheres A and B of radius a and b respectively are at the same potential. The ratio the surface charge density of A and B is
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a) a/b
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b) b/a
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c)a² / b²
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d)b² / a²
Explanation
Potential on surface of sphere=σr / εo Both spheres at same potentialσaa / εo=σbb / εoσa / σb=b/aAnswer: (b)
Q.35
A uniform wire of length 5m is carrying a steady current. The electric field inside it is 0.2 V/m. The potential difference across the ends of wire is
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a) 1.0 volt
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b) 0.5 volt
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c)0.1 volt
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d)5 volt
Explanation
V=E.d=02×5=1V Answer:(a)
Q.36
A charge is divided into parts q1 and (q-q1). What is the ratio q/q1 so that the force between the two parts placed a given distance apart is maximum?
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a) 1:1
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b) 2:1
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c) 1:2
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d) 1:4
Explanation
F is maximum when (q-q1)=q1 or q/q1=2 : 1 Answer: (b)
Q.37
Let P(r)=Qr/ πR4 be the charge density distribution for a solid sphere of radius R and total charge Q. For a point 'p' inside the sphere at distance r1 from the centre of the sphere, the magnitudes of electric field is .. [ AIEEE 2009]
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a)
0%
b)
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c)
0%
d)zero
Explanation
let us consider a spherical shell of thickness dr and radius r. the volume of this shell=4πr2dr. The charge enclosed within shell.The charge enclosed by the sphere of radius r1∴ The electric field at point p inside the sphere at a distance r1 from the centre of the sphere is Answer: (b)
Q.38
A hollow metal sphere of radius 5cms is charged such that the potential on its surface is 10 Volts. The potential at the centre of the sphere is [ IIT 1983]
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a) zero
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b) 10 Volts
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c)same as at a point 5 cm away from the surface
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d)same as at a point 25 cm away from the surface
Explanation
Potential inside the hollow sphere is same at every point and is equal to potential on the surfaceAnswer: (b)
Q.39
Two point charges +q and -q are held fixed at ( -d,0) and (d,0) respectively of a x-y coordinate system. Then [ IIT 1995]
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a)the electric field E at all points on the x-axis has the same direction
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b) electric field at all points on y-axis is along x-axis
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c)Work has to be done in bringing a test charge from ∞ to the origin
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d)the dipole moment is 2qd along x-axis
Explanation
option (a) :electric field at any point on axial line is along the direction of dipole moment which is on negative x-axis Thus incorrectoption (b): If we take any point on Y axis then we find net electric field along +X-direction Thus option is correctoption c: Potential at equatorial ine of dipole is zero, so potential at (0,) is zero. Potential at ∞ is also zero thus no work is done Option (c) is incorrectOption d) The direction of dipole moment is from -ve to +ve. Therefore option is incorrect Answer:(b)
Q.40
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 a potential difference 2V. The charging battery is now disconnected and capacitor are connected in parallel to each other in such a way that the positive terminal of one is connected to the negative terminal. The final energy of the configuration is [ IIT 1995]
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a) zero
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b) (3/2) CV2
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c) (25/6) CV2
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d) (9/2) CV2
Explanation
C and 2C are in parallel to each other ∴ Resultant capacitance=2C+C=3C Net potential=2V - V=V ∴ Final energy=½ CR VR Final energy=½3CV2=(3/2) CV2 Answer: (b)
Q.41
An electron of mass me initially at rest, moves through a certain distance in a uniform electric field in time tA proton of mass mp, also, initially at rest, takes time t2 to move through an equal distance in this uniform electric field. Neglecting the effect of gravity, the ratio t2 / t1 is nearly equal to [ IIT 1997]
0%
a)1
0%
b) ( mp / me)1/2
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c)( me / mp)1/2
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d)1836
Explanation
A charge on proton and electron is same therefore electric force on both will be same but acceleration will be different due to different mass Displacement of electron of x in time t1 acceleration=Ee/ me x=ut + ½ at12 x=½ Ee/ me ×t12 Displacement of x of proton in time t2 acceleration=Ee/ mp x=ut + ½ at22 x=½ Ee/ mp ×t12 ∴ t2 / t1=( mp / me ) 1/2Answer: (b)
Q.42
Two identical metal plates are given positive charge Q 1 and Q2 ( Q2< Q1) respectively. If they are now brought close together to form a parallel plate capacitor with capacitor with capacitance C, the potential difference between them is [ IIT 1999]
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a)
0%
b)
0%
c)
0%
d)
Explanation
Let A be the area of plate and d be the distance between the plates Within capacitor electric field due to individual plate Answer:(d)
Q.43
For the circuit shown in figure. Which of the following statement is true[ IIT 1999]
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a) With S1 closed V1=15V, V2=20V
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b) With S3 closed V1=V2=25V
0%
c) With S1 and S2 closed V1=V2=0
0%
d) With S1 and S3 closed V1=30V, V2=20V
Explanation
With the closing of switch S3 and S1 the negative charge on C2 will attract the positive charge on C1. The negative charge on C1 will attract positive charge on C1. No transfer of charge will take place. Therefore p.d. across C1 and C2 will be 30V and 20 V Answer: (d)
Q.44
if a capacitor C, 3C, 5C ...∞ is one network and 2C, 4C, 6C ...∞ is another network are connected in series. If the effective capacitance of first network is C1 and that of network second be C2 then C1C2 /(C2 -C1) equal to
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a) C / log2
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b) log2/C
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c) ∞
0%
d) (log2)2 / C
Explanation
Answer: (a)
Q.45
Three charges Q, +q and +q are placed at the vertices of a right-angled isosceles triangle as shown in figure. The net electrostatic energy of the configuration ia zero if Q is equal to .. [ IIT 2000]
0%
a)-q / (1+√2)
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b) -2q / (2+√2)
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c)-2q
0%
d)+q
Explanation
we have Answer: (b)
Q.46
A parallel plate capacitor of area A, plate separation d and capacitance C is filled with three different dielectric materials having dielectric constants k1, k2 and k3 as shown. If a single dielectric material is to be used to have the same capacitance C in this capacitor, then its dielectric constant k is given by [ IIT 2000]
0%
a)
0%
b)
0%
c)
0%
d)
Explanation
Let C1 be the capacitance with K1C2 be the capacitance with K2C3 be the capacitance with K3For C1 Area=A/2 and distance=d/2 For C2 Area=A/2 and distance=d/2 For C2 Area=A and distance=d/2 From figure it is clear that C1 and C2 are parallel let it be C' and C' is in series with C3 But C"=εoKA/ d comparing value of K option (b) is correctAnswer: (b)
Q.47
Consider the situation shown in the figure. The capacitor A has a charge q on it whereas B is uncharged. The charge appearing on the capacitor B a long time after the switch is closed is ..[ IIT 2001]
0%
a)zero
0%
b)q/2
0%
c)q
0%
d)2q
Explanation
Since capacitor B is not grounded thus there will not be any transfer of charge and charge on capacitor B will be zero Answer:(a)
Q.48
A uniform electric field pointing in positive x-direction exists in a region. Let A be the origin, B be the point on the x-axis at x=+1 cm and C be the point on the y-axis at y=+1cm. Then the potential at the point A, B, and C satisfy [ IIT 2001]
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a) VA < VB
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b) VA > VB
0%
c) VA < VC
0%
d) VA > VC
Explanation
Electric field is along positive x-direction and dv=-E dx. Thus as we go away on positive direction on x axis potential will reduce thus VA > VB Answer: (b)
Q.49
Two equal point charges are fixed at x=-a and x=+a on the x-axis. Another point charge Q is placed at the origin. The change in the elecrical potential energy of Q, when it is displaced by a small distance x along the x-axis, is approximately proportional to [ IIT 2002]
0%
a)x
0%
b) x2
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c)x3
0%
d)1/x
Explanation
Initial potential energy of Q Final potential energy of Q Change in potential energy of Q=Uf - Ui Thus change in potential energy ∝ x2Answer: (b)
Q.50
Two identical capacitors, have the same capacitance C. One of them is charged to potential V1 and the other VThe negative ends of the capacitor are connected together. When the positive ends are also connected, the decrease in the energy of the combined system is [ IIT 2002]
0%
a)
0%
b)
0%
c)
0%
d)
Explanation
Before connecting Charge on first capacitor q1=CV1Charge on second capacitor=q2=CV2 Total Energy of capacitors before joining After joining total charge will be conserved and capacitance will be 2C final energy Change in energy Uf-Ui Answer: (c)
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