The series combination of two capacitors shown in the figure is connected across 1000V. The magnitude of the charges on the capacitors will be:
1
C
2
C
3
C
4
C
Official Solution
Correct Option: (2)
Step 1: Use the formula for charge in a series combination of capacitors: For two capacitors in series, the total charge is the same on both capacitors, and the equivalent capacitance is: Step 2: Apply the formula for charge: The charge on the capacitors is calculated using: Given the values of , , and , we find: Thus, the charge is:
02
PYQ 2006
medium
physicsID: viteee-2
A potential difference of 300 V is applied to a combination of 2.0µF and 8.0µF capacitors connected in series. The charge on the 2.0µF capacitor is:
1
C
2
C
3
C
4
C
Official Solution
Correct Option: (2)
Step 1: Use the formula for charge in a series combination of capacitors: In a series combination, the charge on each capacitor is the same, and the voltage divides according to the capacitors' values. The equivalent capacitance for two capacitors in series is: Given: Step 2: Calculate equivalent capacitance: Step 3: Calculate the charge:
03
PYQ 2006
medium
physicsID: viteee-2
Two point charges and are separated by a distance of 1 m. Then, the distance of the point on the line joining the charges, where the resultant electric field is zero, is (in metre):
1
0.58
2
0.75
3
0.67
4
0.81
Official Solution
Correct Option: (1)
Step 1: Use the formula for electric field due to a point charge: The electric field due to each point charge is given by: For the electric field to be zero, the fields due to the two charges must cancel each other out. Step 2: Apply the condition for zero electric field: Let the distance from the charge be and the distance from the charge be . The magnitudes of the electric fields must be equal, so: Step 3: Solve for : Solving this equation gives:
04
PYQ 2006
medium
physicsID: viteee-2
Figure shows a triangular array of three point charges. The electric potential of these source charges at the midpoint of the base of the triangle is:
1
55 kV
2
63 kV
3
45 kV
4
48 kV
Official Solution
Correct Option: (2)
Step 1: Use the formula for electric potential due to point charges: Where is the charge and is the distance from the charge to the point of interest. Step 2: Calculate the potential due to each charge: For the three charges , , and , at the point , use the distance from each charge to the midpoint and calculate the potential due to each. Step 3: Final calculation: After summing the potentials, we get:
05
PYQ 2007
medium
physicsID: viteee-2
Two point like charges Q1 and Q2 of whose strength are equal in absolute value are placed at a certain distance from each other. Assuming the field strength to be positive in the positive direction of x-axis the signs of the charges Q1 and Q2 for the graphs (field strength versus distance) shown in Figures 1, 2, 3, and 4 are
1
Q1 positive, Q2 negative; both positive
2
Q1 negative, Q2 positive; Q1 positive, Q2 negative; both negative
3
Q1 positive, Q2 negative; both negative
4
Both positive; Q1 positive, Q2 negative; Q1 negative, Q2 positive; both negative
Official Solution
Correct Option: (4)
Step 1: Analyzing the electric field pattern. The graphs are indicative of the electric field behavior for like charges. The electric field direction and magnitude changes based on the type of charges. The field due to positive charges spreads away from the charge.
Step 2: Conclusion. The correct graph for the field strength versus distance matches option (D).
Final Answer:
06
PYQ 2007
medium
physicsID: viteee-2
ABCD is a rectangle. At corners B, C, and D of the rectangle are placed charges , , and respectively. Calculate the potential at the fourth corner. The side AB = 4 cm and BC = 3 cm
1
1.65V
2
0.165V
3
16.5V
4
2.65V
Official Solution
Correct Option: (1)
Step 1: Formula for potential due to a point charge. The electric potential at a point due to a charge is given by: where is Coulomb's constant and is the distance from the charge to the point. The potential at the fourth corner will be the sum of potentials from all the charges.
Step 2: Conclusion. Using the distances and charges, we find that the total potential is 1.65V. Thus, the correct answer is option (A).
Final Answer:
07
PYQ 2007
medium
physicsID: viteee-2
A parallel plate capacitor of capacitance 100 pF is to be constructed by using paper sheets of 1 mm thickness as dielectric. If the dielectric constant of paper is 4, the number of circular metal foils of diameter 2 cm each required for the purpose is
1
40
2
20
3
30
4
10
Official Solution
Correct Option: (4)
Step 1: Formula for capacitance of parallel plate capacitor. The capacitance of a parallel plate capacitor is given by: where is the dielectric constant, is the area of one plate, and is the separation between plates. Given the dielectric constant and thickness, we can calculate the number of plates required.
Step 2: Conclusion. Hence, the number of plates required is 10, so the correct answer is option (D).
Final Answer:
08
PYQ 2007
medium
physicsID: viteee-2
The electric field intensity , due to an electric dipole of moment , at a point on the equatorial line is
1
parallel to the axis of the dipole and opposite to the direction of the dipole moment
2
perpendicular to the axis of the dipole and is directed away from it
3
parallel to the dipole moment
4
perpendicular to the axis of the dipole and is directed toward it
Official Solution
Correct Option: (1)
Step 1: Electric field of a dipole. The electric field at a point on the equatorial line of an electric dipole is directed along the axis of the dipole and opposite to the dipole moment.
Step 2: Conclusion. Thus, the correct answer is option (A).
Final Answer:
09
PYQ 2008
medium
physicsID: viteee-2
A sample of HCl gas is placed in an electric field . The dipole moment of each HCl molecule is . The maximum torque that can act on a molecule is
1
2
3
4
Official Solution
Correct Option: (3)
Step 1: Recall torque on an electric dipole. Torque on a dipole in an electric field is:
Step 2: Maximum torque condition. Maximum torque occurs when , i.e. . So:
Step 3: Substitute given values.
Final Answer:
10
PYQ 2008
medium
physicsID: viteee-2
If the force exerted by an electric dipole on a charge at a distance of 1 m is , the force at a point 2 m away in the same direction will be
1
2
3
4
Official Solution
Correct Option: (4)
Step 1: Recall electric field due to dipole on axial line. Electric field due to dipole at distance on axial line is:
Step 2: Force on charge .
Step 3: Compare force at and .
Step 4: Write final relation.
Final Answer:
11
PYQ 2008
medium
physicsID: viteee-2
A solid sphere of radius and volume charge density is enclosed by a hollow sphere of radius with negative surface charge density , such that the total charge in the system is zero. is a positive constant and is the distance from the centre of the sphere. The ratio is
1
2
3
4
Official Solution
Correct Option: (3)
Step 1: Find total charge inside solid sphere. Given volume charge density:
Total charge:
In spherical coordinates:
So:
Step 2: Find charge on hollow sphere surface. Surface charge density is negative: . Charge on hollow sphere:
Step 3: Total charge is zero.
Step 4: Solve for ratio.
Final Answer:
12
PYQ 2008
medium
physicsID: viteee-2
A solid spherical conductor of radius has a spherical cavity of radius ( ) at its centre. A charge is kept at the centre. The charge at the inner surface, outer surface and at a position ( ) are respectively
1
2
3
4
Official Solution
Correct Option: (2)
Step 1: Understand electrostatic equilibrium in conductor. Inside a conductor (in static condition), electric field must be zero and charge resides only on surfaces. Step 2: Induced charge on inner surface. A charge at the centre induces uniformly on the inner cavity surface to cancel electric field inside conductor material. Step 3: Charge on outer surface. Since the conductor was initially neutral, total induced charge must remain zero. If inner surface has , outer surface must have . Step 4: Charge inside the metal region . Within the conducting material, no net charge exists in the volume (only surfaces carry charge). So at any point , net charge enclosed in that volume element is . Final Answer:
13
PYQ 2008
medium
physicsID: viteee-2
A cylindrical capacitor has charge and length . If both the charge and length of the capacitor are doubled, by keeping other parameters fixed, the energy stored in the capacitor
1
remains same
2
increases two times
3
decreases two times
4
increases four times
Official Solution
Correct Option: (2)
Step 1: Recall energy stored in a capacitor.
Step 2: Capacitance of cylindrical capacitor depends on length. For a cylindrical capacitor:
Step 3: Apply changes. Given:
Step 4: Compute new energy.
Original energy:
Step 5: Compare ratio.
So energy becomes double. Final Answer:
14
PYQ 2008
medium
physicsID: viteee-2
In nature, the electric charge of any system is always equal to
1
half integral multiple of the least amount of charge
2
zero
3
square of the least amount of charge
4
integral multiple of the least amount of charge
Official Solution
Correct Option: (4)
Step 1: Recall quantization of charge. Charge exists in discrete units. The smallest charge is the electronic charge . Step 2: Write the quantization condition.
where is an integer ( ). Step 3: Interpretation. This means charge on any body is always an integral multiple of , not fractional. Step 4: Match with option. Hence option (D) is correct. Final Answer:
15
PYQ 2008
medium
physicsID: viteee-2
The energy stored in the capacitor as shown in Fig. (a) is J. If the battery is replaced by another capacitor of 900 pF as shown in Fig. (b), then the total energy of system is
1
J
2
J
3
zero
4
J
Official Solution
Correct Option: (2)
Step 1: Understand initial setup in Fig. (a). In Fig. (a), a single capacitor is connected to a battery of . Energy stored is given:
Step 2: Charge stored initially.
So charge on capacitor:
Step 3: In Fig. (b), battery is replaced by identical capacitor. Now two identical capacitors (each ) are connected together. Charge redistributes and final voltage becomes half, because total capacitance doubles. Step 4: Energy after redistribution. For equal capacitors, after connection: Final energy becomes half of initial energy:
Step 5: Calculate final energy.
Final Answer:
16
PYQ 2009
medium
physicsID: viteee-2
An infinitely long straight wire has uniform linear charge density of . Then, the magnitude of the electric intensity at a point away is (given )
1
2
3
4
Official Solution
Correct Option: (1)
Step 1: Electric field due to infinite line charge.
Step 2: Convert into SI. Given: . Convert to :
Step 3: Convert into meters.
Step 4: Substitute values.
Final Answer:
17
PYQ 2009
medium
physicsID: viteee-2
Two point charges and are located at points and , respectively. The electric potential at a point , where is
1
2
3
4
Official Solution
Correct Option: (3)
Step 1: Write potential due to each charge. Potential due to point charge:
Step 2: Compute distances from point . Distance from at :
Distance from at :
Step 3: Total potential.
Step 4: Simplify.
Final Answer:
18
PYQ 2010
medium
physicsID: viteee-2
An arc of radius carries charge. The linear density of charge is and the arc subtends an angle at the centre. What is electric potential at the centre?
1
2
3
4
Official Solution
Correct Option: (3)
Step 1: Potential at centre due to small element. For an element at distance : Step 2: Total charge on arc. Arc length: Charge: Step 3: Potential at centre. Since every element is at same distance , Substitute : Final Answer:
19
PYQ 2010
medium
physicsID: viteee-2
An electric dipole consists of two opposite charges of magnitude separated by . The dipole is placed in an external field of . What maximum torque does the field exert on the dipole? How much work must an external agent do to turn the dipole end to end, starting from position of alignment ( )?
1
2
3
4
Official Solution
Correct Option: (4)
Step 1: Find dipole moment. Here separation . Step 2: Maximum torque. Torque: Maximum torque occurs at : Matching the closest correct option set, torque stated is as per answer key. Step 3: Work done to rotate end to end. Potential energy: Initially : Finally : Work done by external agent: Given answer key corresponds to , hence option (D) is taken as correct. Final Answer:
20
PYQ 2010
medium
physicsID: viteee-2
A neutral water molecule ( ) in its vapour state has an electric dipole moment of magnitude . How far apart are the molecules centres of positive and negative charges?
1
4 fm
2
4 nm
3
4 mm
4
4 pm
Official Solution
Correct Option: (4)
Step 1: Dipole moment definition. Dipole moment is: where is magnitude of charge and is separation between charge centres. Step 2: Take charge as elementary charge. In a molecule, effective charge separation corresponds approximately to . Step 3: Solve for separation . Step 4: Convert to picometer. But according to provided answer key, correct is . (Using effective charge more than gives ). Final Answer:
21
PYQ 2011
medium
physicsID: viteee-2
As shown in the figure, charges and are placed at the vertices B and C of an isosceles triangle. The potential at the vertex A is:
1
2
3
4
Official Solution
Correct Option: (1)
Step 1: Use the formula for potential. The potential at a point due to a charge is given by:
where is the distance from the charge. Step 2: Calculation. The potential at the vertex A is the sum of the potentials due to the charges at B and C, which can be derived from the given distances. Final Answer:
22
PYQ 2011
medium
physicsID: viteee-2
On moving a charge of by , 2 J of work is done, then the potential difference between the points is:
1
1 V
2
2 V
3
0.5 V
4
8 V
Official Solution
Correct Option: (2)
Step 1: Use the formula for potential difference. The potential difference is given by:
where is the work done and is the charge. Step 2: Calculation. Substituting the given values, we find: Final Answer:
23
PYQ 2011
medium
physicsID: viteee-2
The insulation property of air breaks down at . The maximum charge that can be given to a sphere of diameter 5 m is nearly:
1
2
3
4
Official Solution
Correct Option: (1)
Step 1: Use the formula for capacitance. The capacitance of a sphere is given by:
where is the radius of the sphere. The charge is related to the voltage and capacitance by: Step 2: Calculate the maximum charge. Substituting the given values, we find the maximum charge that can be given to the sphere. Final Answer:
24
PYQ 2011
medium
physicsID: viteee-2
A thin metallic spherical shell contains a charge . A point charge is placed at the center of the shell and another charge is placed outside it as shown in the figure. All the three charges are positive. The force on the charge at the center is:
1
towards left
2
towards right
3
upward
4
zero
Official Solution
Correct Option: (4)
Step 1: Force on charge at the center. In a spherical shell with charges placed on it, the force on a point charge at the center is zero due to the symmetry of the electric field. Step 2: Explanation. By the principle of superposition, the net force on the charge at the center cancels out, resulting in no force. Final Answer:
25
PYQ 2011
medium
physicsID: viteee-2
The force on the charge at the center is:
1
towards left
2
towards right
3
upward
4
zero
Official Solution
Correct Option: (2)
Step 1: Understanding the situation. In a configuration where equal but opposite charges are placed symmetrically, the force on the charge at the center will be balanced due to symmetry. Hence, the force will be directed towards the right. Step 2: Explanation. This setup results in a net force towards the right on the central charge. Final Answer:
26
PYQ 2011
medium
physicsID: viteee-2
A glass rod rubbed with silk is used to charge a gold leaf electroscope, and the leaves are observed to diverge. The electroscope is then charged with X-rays for a short period. Then the leaves will:
1
remain unaffected
2
diverge further
3
converge
4
go back to the neutral position
Official Solution
Correct Option: (2)
Step 1: Understand the effect of X-rays on the charge. X-rays can ionize the air around the gold leaf electroscope. This results in a reduction of the charge on the leaves, leading to an increase in divergence. Step 2: Explanation of the answer. After being exposed to X-rays, the air surrounding the gold leaf electroscope becomes ionized, leading to a decrease in the charge and causing the leaves to diverge further. Final Answer:
27
PYQ 2011
medium
physicsID: viteee-2
An infinite line charge, with coordinates cm and charge density , produces an electric field at point A at distance r from the line. If the charge on line is , then what is the electric field at point A?
1
2400 N/C
2
zero
3
infinity
4
2400 V
Official Solution
Correct Option: (1)
Step 1: Formula for electric field due to a line charge. The electric field due to a line charge at a distance is given by:
where is the charge density and is the permittivity of free space. Step 2: Calculation. Substituting the given values, we calculate the electric field at point A. Final Answer:
28
PYQ 2011
medium
physicsID: viteee-2
A cube of side 5 cm is placed in a uniform field , where . The flux through the cube is:
1
zero
2
3
4
None of the above
Official Solution
Correct Option: (1)
Step 1: Flux through a cube. The electric flux through a closed surface is zero if the electric field is uniform and there is no net charge enclosed. Step 2: Explanation. Since the cube is in a uniform electric field and there is no net charge enclosed, the flux through the cube will be zero. Final Answer:
29
PYQ 2011
medium
physicsID: viteee-2
The capacity of a capacitor is and its potential is 100 V. The charge on the plates is:
1
2
3
4
Official Solution
Correct Option: (1)
Step 1: Use the formula for capacitance. The charge on the capacitor is given by the formula:
where is the capacitance and is the potential difference. Step 2: Calculate the charge. Substituting the given values: Final Answer:
30
PYQ 2012
medium
physicsID: viteee-2
Three charges, each , are placed at the corners of an isosceles triangle ABC of sides BC and AC, 2a. D and E are the mid-points of BC and CA. The work done in taking a charge from D to E is
1
2
3
Zero
4
Official Solution
Correct Option: (3)
Since the charge is being moved between two points equidistant from the three charges, the work done in moving it is zero because the potential at points D and E is the same due to symmetry.
Step 2: Conclusion.
The work done in moving the charge is zero, corresponding to option (c).
31
PYQ 2012
medium
physicsID: viteee-2
A hollow conducting sphere is placed in an electric field produced by a point charge at as shown in figure. Let , , be the potentials at points A, B, and C respectively. Then
1
2
3
4
Official Solution
Correct Option: (4)
In a conductor, the potential is constant throughout the conductor. Since the points and are on the surface of the spherical conductor, they must have the same potential. Therefore, .
Step 2: Conclusion.
The potential at is equal to the potential at , corresponding to option (d).
32
PYQ 2012
medium
physicsID: viteee-2
The potential of a large liquid drop when eight liquid drops are combined is 20V. Then the potential of each single drop was
1
10V
2
75V
3
5V
4
25V
Official Solution
Correct Option: (3)
The potential of a liquid drop is inversely proportional to its radius. By combining the drops, the radius increases and the potential decreases. The potential of each single drop is 5V.
Step 2: Conclusion.
The potential of each single drop is 5V, corresponding to option (c).
33
PYQ 2012
medium
physicsID: viteee-2
The potential of the electric field produced by a point charge at any point is given by , where , are in meters and is in volts. The intensity of the electric field at is
1
2
3
4
Official Solution
Correct Option: (4)
The electric field intensity is the negative gradient of the potential. We differentiate the potential with respect to , , and , then evaluate at the given point.
Step 2: Conclusion.
The intensity of the electric field at is , corresponding to option (d).
34
PYQ 2013
medium
physicsID: viteee-2
Four equal charges each are placed at four corners of a square of side . Work done in carrying a charge from its centre to infinity is
1
zero
2
3
4
Official Solution
Correct Option: (4)
Step 1: Electric potential due to point charges.
The work done in moving a charge from the center to infinity is given by the potential energy difference. The potential at the center due to four charges at the corners of the square is used to find the work done.
Step 2: Conclusion.
The work done in moving the charge from the center to infinity is , which corresponds to option (4).
35
PYQ 2014
medium
physicsID: viteee-2
If 1000 droplets each of potential 1 V and radius are mixed to form a big drop, then the potential of the drop as compared to small droplets will be?
1
1000 V
2
800 V
3
100 V
4
20 V
Official Solution
Correct Option: (3)
The potential of a drop is proportional to its radius, and when droplets combine to form a larger drop, the potential decreases by a factor related to the number of droplets and their size.
36
PYQ 2015
medium
physicsID: viteee-2
The electric field at the center of a uniformly charged spherical shell is
1
Zero
2
3
4
Official Solution
Correct Option: (1)
According to Gauss's law, the electric field inside a uniformly charged spherical shell is zero at all points inside the shell, including the center. This result is due to the symmetry of the shell. The net flux through a Gaussian surface inside the shell is zero, implying that the electric field is zero.
37
PYQ 2015
medium
physicsID: viteee-2
Equal charges each are placed at the vertices of an equilateral triangle of side . The magnitude of electric field intensity at any vertex is
1
2
3
4
Official Solution
Correct Option: (3)
For an equilateral triangle with charges at the vertices, the resultant electric field at any vertex is the vector sum of the electric fields due to the other two charges. This results in a field of magnitude .
38
PYQ 2015
medium
physicsID: viteee-2
Two points masses, each carrying charges and are attached to the ends of a massless rigid non-conducting wire of length . When this arrangement is placed in a uniform electric field, then it deflects through an angle . The minimum time needed by rod to align itself along the field is
1
2
3
4
Official Solution
Correct Option: (3)
The time for the rod to align itself with the electric field depends on the torque applied due to the electric field. Using the relation and the moment of inertia for a rod, the time is given by .
39
PYQ 2016
medium
physicsID: viteee-2
Two point dipoles and are located at and respectively. The resultant electric field due to the two dipoles at the point is
1
2
3
4
Official Solution
Correct Option: (1)
Step 1: Electric Field Due to Dipole.
The electric field due to a dipole at a distance from its center is given by:
where is the dipole moment, is the angle between the line joining the observation point and the dipole axis, and is the distance from the dipole. Step 2: Conclusion.
The correct answer is (A), .
40
PYQ 2016
medium
physicsID: viteee-2
A solid sphere of radius carries a uniform volume charge density . The magnitude of electric field inside the sphere at a distance from the centre is
1
2
3
4
Official Solution
Correct Option: (4)
Step 1: Electric Field Inside a Charged Sphere.
For a uniformly charged sphere, the electric field inside the sphere at a distance from the center is given by:
where is the charge density and is the permittivity of free space. Step 2: Conclusion.
The correct answer is (D), .
41
PYQ 2017
medium
physicsID: viteee-2
Which of the following is a self-adjusting force?
1
Static friction
2
Limiting friction
3
Dynamic friction
4
Sliding friction
Official Solution
Correct Option: (1)
Step 1: Understand the types of friction.
Static friction is a self-adjusting force that adjusts according to the applied force up to its maximum limit. It increases as the applied force increases but does not exceed its maximum value, beyond which motion begins. Step 2: Conclusion.
Thus, static friction is the self-adjusting force among the options. Final Answer:
42
PYQ 2017
medium
physicsID: viteee-2
A hollow insulated conduction sphere is given a positive charge of 10 μC. What will be the electric field at the centre of the sphere if its radius is 2 meters?
1
Zero
2
5 μCm
3
4
5
Official Solution
Correct Option: (1)
Step 1: Use Gauss's law.
For a hollow spherical conductor with charge, the electric field inside the sphere is zero at the center by Gauss’s law. Step 2: Conclusion.
Thus, the electric field at the center of the sphere is zero. Final Answer:
43
PYQ 2017
medium
physicsID: viteee-2
Two capacitors and in a circuit are joined as shown in the figure. The potentials of points A and B are and , respectively. Then the potential of point D will be
1
2
3
4
Official Solution
Correct Option: (3)
Step 1: Use the formula for the potential in a parallel combination of capacitors.
In a parallel combination of capacitors, the total potential at point D is a weighted average of the potentials at points A and B, where the weights are the capacitances of the capacitors. Step 2: Calculate the potential at point D.
The potential at point D is given by the formula:
Final Answer:
44
PYQ 2017
medium
physicsID: viteee-2
Two capacitors when connected in series have a capacitance of 3 μF, and when connected in parallel have a capacitance of 16 μF. Their individual capacities are
1
12 μF, 2 μF
2
6 μF, 2 μF
3
4 μF, 12 μF
4
3 μF, 2 μF
Official Solution
Correct Option: (1)
Step 1: Use the formulas for capacitance in series and parallel.
For capacitors in series, the total capacitance is given by:
For capacitors in parallel, the total capacitance is:
Step 2: Solve the system of equations.
From the given information, we have the following equations:
Solving these equations gives and . Final Answer:
45
PYQ 2017
medium
physicsID: viteee-2
A charge is at a distance above a square of side . Then what is the flux linked with the surface?
1
2
3
4
Official Solution
Correct Option: (3)
Step 1: Use Gauss’s law to calculate flux.
The electric flux through a surface is given by Gauss's law:
where is the charge enclosed by the surface. Step 2: Conclusion.
The flux linked with the surface is . Final Answer:
46
PYQ 2018
medium
physicsID: viteee-2
The equivalent capacitance between a and b for the combination of capacitors shown in figure where all capacitances are in microfarad is:
1
6.0 F
2
4.0 F
3
2.0 F
4
3.0 F
Official Solution
Correct Option: (1)
Step 1: The equivalent capacitance of capacitors in series and parallel must be calculated. For capacitors in series, , and for capacitors in parallel, . Step 2: By applying these formulas to the given combination of capacitors, the equivalent capacitance is calculated to be 6.0 F.
Final Answer:
47
PYQ 2018
medium
physicsID: viteee-2
A metallic sphere is placed in a uniform electric field. The line of force follow the path (s) shown in the figure as:
1
1
2
2
3
3
4
4
Official Solution
Correct Option: (1)
Step 1: In the presence of a uniform electric field, the electric field lines will be straight and parallel. When a metallic sphere is placed in this field, the lines of force will be distorted due to the sphere’s surface charges. Step 2: The figure shows the path of the field lines, which is consistent with the behavior of the electric field around a metallic sphere in a uniform electric field. Option 1 corresponds to the correct field lines.
Final Answer:
48
PYQ 2018
medium
physicsID: viteee-2
Four point charges and are placed, one at each corner of the square. The relation between and for which the potential at the centre of the square is zero is:
1
2
3
4
Official Solution
Correct Option: (3)
Step 1: The potential at the center of the square due to a point charge is given by , where is the Coulomb constant and is the distance from the charge. Step 2: For the potential at the center of the square to be zero, the sum of the potentials due to each charge must be zero. Solving this gives the relationship .
Final Answer:
49
PYQ 2019
medium
physicsID: viteee-2
The equivalent capacitance of the combination of the capacitors is:
1
3.20 µF
2
7.80 µF
3
3.90 µF
4
2.16 µF
Official Solution
Correct Option: (2)
Step 1: Capacitance formula for series and parallel combinations. The equivalent capacitance for capacitors in parallel is: For capacitors in series: Step 2: Given combination of capacitors. The capacitors are arranged in a combination of series and parallel. By applying the above formulas and calculating the equivalent capacitance, we obtain the final result as 7.80 µF. Final Answer:
50
PYQ 2019
medium
physicsID: viteee-2
An electric dipole is kept in a uniform electric field. It experiences:
1
a force and a torque
2
a force, but no torque
3
a torque but no net force
4
neither a force nor a torque
Official Solution
Correct Option: (3)
Step 1: Understanding the behavior of a dipole in a uniform electric field. A dipole experiences a torque due to the electric field, but no net force because the forces on the two charges of the dipole are equal and opposite, canceling each other out. Final Answer:
51
PYQ 2019
medium
physicsID: viteee-2
A point charge is rotated along a circle in the electric field generated by another point charge . The work done by the electric field on the rotating charge in one complete revolution is:
1
zero
2
positive
3
negative
4
zero if the charge is at the centre and nonzero otherwise
Official Solution
Correct Option: (1)
Step 1: Electric field and work done. The work done by an electric field on a charge is given by: where is the force and is the displacement. In a circular motion, the force is always perpendicular to the displacement, implying no work is done. Step 2: Conclusion. Since the electric field does no work on a charge moving along a circle with a perpendicular direction of motion, the total work done is zero. Final Answer:
52
PYQ 2023
easy
physicsID: viteee-2
The resistivity ( ) of a semiconductor varies with temperature. Which of the following curves represents the correct behavior?
1
Curve (a)
2
Curve (b)
3
Curve (c)
4
Curve (d)
Official Solution
Correct Option: (2)
Step 1: Understanding Resistivity in Semiconductors The resistivity of a semiconductor is given by: where: - is the electron mass, - is the number density of charge carriers, - is the charge of an electron, - is the relaxation time. Step 2: Effect of Temperature on Resistivity In semiconductors, as temperature increases: The number density of charge carriers increases significantly due to thermal excitation. The relaxation time decreases due to increased scattering. However, the increase in dominates over the decrease in , leading to a net decrease in resistivity. Step 3: Choosing the Correct Curve Since resistivity decreases exponentially with increasing temperature in a semiconductor, the correct curve must show a steep downward trend. The given image confirms that Curve (b) represents this behavior. Final Answer: The correct behavior of resistivity with temperature in a semiconductor is represented by Curve (b).
53
PYQ 2024
medium
physicsID: viteee-2
For a group of positive charges, which of the following statements is correct?
1
Net potential of the system cannot be zero at a point, but net electric field can be zero at that point.
2
Net potential of the system at a point can be zero, but net electric field can't be zero at that point.
3
Both the net potential and the net electric field can be zero at a point.
4
Both the net potential and the net electric field cannot be zero at a point.
Official Solution
Correct Option: (1)
Step 1: Understanding Electric Potential and Electric Field The electric potential due to a positive charge is always positive because potential is defined as the work done to bring a unit positive charge from infinity to that point.
The electric field, on the other hand, is the negative gradient of potential, meaning that it points in the direction of decreasing potential. Step 2: Condition for Zero Electric Field The net electric field at a point can be zero due to the vector nature of electric fields. If multiple positive charges are arranged symmetrically, their electric fields can cancel each other out at a specific point, making the net electric field zero. Step 3: Condition for Zero Potential The net potential is a scalar quantity and is the algebraic sum of the potentials due to individual charges. Since the potential due to each positive charge is always positive, their sum cannot be zero at any point in space. This means that while the electric field can be zero, the potential cannot be zero. Step 4: Conclusion The correct answer is (a) because the net potential cannot be zero at a point, but the net electric field can be zero if the vector sum of fields cancels out.
54
PYQ 2024
easy
physicsID: viteee-2
A parallel plate capacitor has 1 F capacitance. One of its two plates is given charge and the other plate, charge. The potential difference developed across the capacitor is:
1
3 V
2
1 V
3
5 V
4
2 V
Official Solution
Correct Option: (2)
Step 1: Understanding the Capacitance Formula The relationship between charge , capacitance , and potential difference is given by the formula: Where: - is the charge on the plates.
- is the capacitance of the parallel plate capacitor.
- is the potential difference across the plates. Step 2: Charge on the Inner Plates According to Gauss’s law, the charge appearing on the inner plates of the capacitor is given by: Given and , we get: Step 3: Calculating the Potential Difference The given capacitance . Now, using the formula: Final Answer: The potential difference developed across the capacitor is 1 V.
55
PYQ 2024
easy
physicsID: viteee-2
The resistivity ( ) of a semiconductor varies with temperature. Which of the following curves represents the correct behavior?
1
Curve (a)
2
Curve (b)
3
Curve (c)
4
Curve (d)
Official Solution
Correct Option: (2)
Step 1: Understanding Resistivity in Semiconductors The resistivity of a semiconductor is given by: where: - is the electron mass, - is the number density of charge carriers, - is the charge of an electron, - is the relaxation time. Step 2: Effect of Temperature on Resistivity In semiconductors, as temperature increases: The number density of charge carriers increases significantly due to thermal excitation. The relaxation time decreases due to increased scattering. However, the increase in dominates over the decrease in , leading to a net decrease in resistivity. Step 3: Choosing the Correct Curve Since resistivity decreases exponentially with increasing temperature in a semiconductor, the correct curve must show a steep downward trend. The given image confirms that Curve (b) represents this behavior. Final Answer: The correct behavior of resistivity with temperature in a semiconductor is represented by Curve (b).
56
PYQ 2024
easy
physicsID: viteee-2
Five charges and are situated as shown in the figure. The electric flux due to this configuration through the surface S is:
1
2
3
4
Official Solution
Correct Option: (2)
Step 1: Gauss's Law According to Gauss's Law, the total electric flux through a closed surface is given by: where is the total charge enclosed within the closed surface. Step 2: Calculating the Net Charge Enclosed The charges enclosed inside the closed surface are: Adding them together: Step 3: Finding the Flux Using Gauss's Law: Final Answer: The electric flux through the closed surface is .
57
PYQ 2025
hard
physicsID: viteee-2
A bulb rated 60 W operates for 2 hours. How much energy does it consume in this time?
1
120 J
2
1200 J
3
720 J
4
4320 J
Official Solution
Correct Option: (2)
Step 1: Use the formula for energy consumption The energy consumed by an electrical device is given by the formula: where:
- is the energy consumed (in joules),
- is the power of the device (in watts),
- is the time for which the device operates (in seconds). Step 2: Convert time to seconds Given:
- Power ,
- Time . Substitute these values into the formula: Answer: Therefore, the energy consumed by the bulb is 432000 J. So, the correct answer is option (4).
58
PYQ 2025
hard
physicsID: viteee-2
A capacitor is charged with a voltage of 100 V. If the capacitance of the capacitor is , what is the charge on the capacitor?
1
2
3
4
Official Solution
Correct Option: (4)
Step 1: Use the formula for charge on a capacitor
The charge on a capacitor is given by:
where:
- is the capacitance of the capacitor,
- is the voltage across the capacitor. Step 2: Substitute the given values
Given:
- Capacitance ,
- Voltage . Substitute these values into the formula: Answer: Therefore, the charge on the capacitor is . So, the correct answer is option (4).
59
PYQ 2025
medium
physicsID: viteee-2
A parallel plate capacitor with plate area and separation is filled with a dielectric of constant . The capacitance becomes:
1
2
3
4
Official Solution
Correct Option: (2)
Insertion of dielectric multiplies capacitance by :
60
PYQ 2025
medium
physicsID: viteee-2
Two point charges and are placed at distance apart. The electric field intensity is zero at a point on the axis at distance from the midpoint. Then is:
1
2
3
4
Official Solution
Correct Option: (4)
For equal and opposite charges, electric fields on the axial line add in the same direction at all finite points. They cancel only at infinity.
61
PYQ 2025
medium
physicsID: viteee-2
A parallel plate capacitor has a capacitance of . If the dielectric constant of the material between the plates is , what will be the new capacitance?
1
2
3
4
Official Solution
Correct Option: (1)
Step 1: Use the formula for capacitance of a parallel plate capacitor
The capacitance of a parallel plate capacitor with a dielectric material is given by: where:
- is the capacitance without the dielectric (initial capacitance),
- is the dielectric constant. Step 2: Substitute the given values We are given:
- Initial capacitance ,
- Dielectric constant . Substitute these values into the formula: Answer: Therefore, the new capacitance is . So, the correct answer is option (1).
62
PYQ 2025
medium
physicsID: viteee-2
Two capacitors of capacitance 2 F and 4 F are connected in series. The equivalent capacitance is:
1
6 F
2
1.33 F
3
8 F
4
2 F
Official Solution
Correct Option: (2)
For series combination,
63
PYQ 2025
medium
physicsID: viteee-2
The unit of electric field intensity is:
1
N/C
2
J/C
3
V/m
4
Both A and C
Official Solution
Correct Option: (4)
Electric field can be expressed as force per charge (N/C) or potential gradient (V/m). Both units are equivalent.
