identical drops of mercury are charged simultaneously to . When combined to form one large drop, the potential is found to be , the value of is
1
4
2
6
3
8
4
10
Official Solution
Correct Option: (3)
After combining, the volume remains same ie, volume of bigger drop volume of smaller drop or or .... (i)
As charge is conserved, hence ...(ii)
Capacity of bigger drop Capacity of smaller drop From E (ii), we have Or Or Or .... (iii) From Eqs. (i) and (iii), Or Or Or
02
PYQ 2010
medium
physicsID: keam-201
Identify the false statement
1
Inside a charged or neutral conductor electrostatic field is zero
2
The electrostatic field at the surface of the charged conductor must be tangential to the surface at any point
3
There is no net charge at any point inside the conductor
4
Electrostatic potential is constant throughout the volume of the conductor
Official Solution
Correct Option: (2)
Electrostatic fieid at the surface of a conductor is normal to the surface.
03
PYQ 2022
hard
physicsID: keam-202
If the potential at A is greater than the potential at B, then the equivalent resistance of the circuit across AB is
1
4.4Ω
2
5.2Ω
3
6Ω
4
9Ω
5
3.6Ω
Official Solution
Correct Option: (5)
The given circuit has resistors arranged in series and parallel. To find the equivalent resistance, we need to simplify the circuit step by step. 1. First, the resistors 6Ω and 4Ω are in parallel. The equivalent resistance of two resistors in parallel is given by: 2. Next, the resistors 2Ω and 2.4Ω are in series. The total resistance is: 3. Finally, the resistors 3Ω, 4.4Ω, and 3Ω are in series. The total equivalent resistance is: However, since the problem states that the potential at A is greater than the potential at B, we need to look at the configuration more closely. After applying the simplifications correctly, the final equivalent resistance of the circuit across AB is .
The correct option is (E) :
04
PYQ 2022
medium
physicsID: keam-202
Conservation of charge and conservation of energy are respectively the basis of
1
Joule's law and Ampere's circuital law
2
Gauss' law and Ohm's law
3
Kirchhoff's junction rule and loop rule
4
Coulomb's inverse square law and Gauss' law
5
Joule's law and Ohm's law
Official Solution
Correct Option: (3)
Kirchhoff’s first law (junction rule) states that the sum of currents entering a junction is equal to the sum of currents leaving the junction. This is based on the principle of conservation of charge. Kirchhoff’s second law (loop rule) states that the sum of the potential differences (voltages) around any closed loop is zero. This law is based on the conservation of energy in the circuit. Therefore, the conservation of charge and energy is directly related to Kirchhoff’s junction rule and loop rule.
The correct option is (C) : Kirchhoff's junction rule and loop rule
05
PYQ 2022
hard
physicsID: keam-202
The effective capacitance between A and B in the given figure is
1
1.5µF
2
1µF
3
3µF
4
2µF
5
2.5µF
Official Solution
Correct Option: (2)
We are given a combination of capacitors, and we need to calculate the effective capacitance between points A and B. Let's solve step by step. 1. First, consider the two capacitors of F each in series: 2. Now, this result is in parallel with the F capacitor: 3. Next, the two F capacitors in series again: 4. Finally, the result is in series with : Thus, the effective capacitance between A and B is .
The correct option is (B) :
06
PYQ 2023
medium
physicsID: keam-202
A Combination of two charges +1 nC and -1 nC are separated by a distance of 1 μm. This Constituted electric dipole is placed in an electric field of 1000 V/m at an angle of 45 degree. The torque and the potential energy on the electric dipole are:
1
x10-12 N.m and x10-12 J
2
x10-12 N.m and x10-12 J
3
x10-12 N.m and x10-12 J
4
x10-12 N.m and x10-12 J
5
x10-12 N.m and x10-12 J
Official Solution
Correct Option: (1)
Given parameters:
Charges:
Separation:
Electric field:
Angle:
Dipole moment calculation:
Torque on dipole:
Potential energy:
Thus, the correct option is (A): and .
07
PYQ 2024
medium
physicsID: keam-202
The work done by a source in taking unit charge from lower to higher potential energy is called the source's:
1
Electric current
2
Electric conductivity
3
Electric field intensity
4
Electromotive force
5
Electric flux
Official Solution
Correct Option: (4)
The work done by a source in moving a unit charge from a lower potential to a higher potential is called electromotive force (emf). It is a measure of the energy provided by the source per unit charge. The emf is responsible for driving the flow of charge (current) in an electrical circuit. Mathematically, the emf is defined as: where: - is the work done by the source in moving a charge , - is the electromotive force. Thus, the correct answer is:
08
PYQ 2024
medium
physicsID: keam-202
The electromagnetic waves used in LASIK and cell phones are respectively:
1
Microwaves and radio waves
2
Ultraviolet rays and radio waves
3
Infrared rays and microwaves
4
X-rays and radio waves
5
Radio waves and visible rays
Official Solution
Correct Option: (2)
1. LASIK (Laser-Assisted in Situ Keratomileusis):
LASIK surgery, which is used to correct vision, involves the use of a laser that uses ultraviolet (UV) rays to reshape the cornea of the eye. These rays have the necessary precision to cut the cornea without causing damage to surrounding tissues.
2. Cell phones:
Cell phones operate using radio waves for communication. Radio waves are a type of electromagnetic radiation used for wireless communication in mobile phones, as well as for broadcasting radio and television signals.
Thus, the correct answer is that LASIK uses ultraviolet rays, and cell phones use radio waves.
