250+ TOP MCQs on Electrostatic Potential | Class12 Physics

Physics Multiple Choice Questions on “Electrostatic Potential”.

1. Work done to bring a unit positive charge from infinity to a point in an electric field is known as _______
a) Electric potential
b) Electric field intensity
c) Electric dipole moment
d) The total energy of the point charge

Answer: a
Clarification: Electric potential is defined as the amount of work done to bring a unit positive charge from an infinite distance to a particular point of an electric field. The total energy of that point charge means the sum of kinetic energy and potential energy which is not the same as the potential energy if the particle is in motion.

2. If a charged body is moved in an electric field against the Coulomb force, then ________
a) Work is done on the body by an external agent
b) Work is done by the electric field
c) Electric field intensity decreases
d) The total energy of the system decreases

Answer: a
Clarification: To move a body against some force, work is to be done on the body. In this case, an external force is to be applied on the body to move it i.e. an external work is to be done. As we are moving the body against the Coulomb’s force, hence no work is done on the body by the electric field.

3. What is the dimension of electric potential?
a) [M L T-2]
b) [M L T-3 I]
c) [M L T-3 I-1]
d) [M L2 T-3 I]

Answer: c
Clarification: Potential can be simply defined as work done on a unit charge, therefore the dimension=(frac {the , dimension , of , work}{the , dimension , of , charge}=frac {[M L T^{-2}]}{[I T]})=[M L T-3 I-1]. The dimension of potential energy in mechanics is not the same as electric potential energy though both of them are the unit of energy. In mechanics, it is defined as work done on a unit mass.

4. Electric potential varies with distance such that V(x) =ax-bx3; where a and b are constants. Where will the electric field intensity be zero?
a) x=(frac {a}{b})
b) x=(frac {a}{3b})
c) x=(sqrt{frac {a}{b}})
d) x=+(sqrt{frac {a}{3b}}) and x=-(sqrt{frac {a}{3b}})

Answer: d
Clarification: We know that E=-(frac {dv}{dx})
∴ E = –(frac {d}{dx}) (ax – bx3) = -(a – 3bx2)
Electric field will be zero. ∴ E = 0 ⇒ (a-3bx2) = 0
⇒ a = 3bx2
⇒ x2 = (frac {a}{3b}) = x = ±(sqrt{frac {a}{3b}})

5. 1 electron volt= __________ J.
a) 1.6*10-19
b) 4.8*10-19
c) 1.6*10-10
d) 10

Answer: a
Clarification: 1 electron volt is the amount of work done if an electron is passed through a potential difference of 1V. Therefore the work done = 1V*charge of an electron = 1.602*10-19 J. But it is a small quantity and hence we use kilo electron volt and mega electron volt in practical.

6. 1 V/m= _______
a) 1N/C
b) 3*1010 N/C
c) 107 N/C
d) 1010 N/C

Answer: a
Clarification: From the definition, we know that electric field E=(frac {-dv}{dx}). Therefore V/m is the unit of electric field intensity. 1 V/m means the amount of electric field in which if we move a unit positive charge by 1 m, the work done will be 1N. Therefore, the electric field will be 1N/C.

7. The electric potential difference between two points is a path function. The statement is __________
a) True
b) False

Answer: b
Clarification: Electric potential depends only on the electric field intensity and the amount of charge. It has no dependency on the path by which the charge is traveling. Therefore, if we move a charge from one point to another in presence of an electric field in a straight line or a curve line, work done in both the cases will be the same i.e. potential difference between the points will be the same.

8. By performing a set of experiments, a scientist found that the electric field between two points A and B is zero. What can he conclude regarding the potential of the two points?
a) VA=0, VB=0
b) VA>VB
c) VA+VB
d) VA=VB

Answer: d
Clarification: We know thatVAB=VA-VB=-(int_A^B)E.dx. But E=0, so the value of the integral becomes zero. Therefore VA=VB is obtained. The electric field is always directed from a point of higher potential to a point of lower potential. But if the potential of two points is the same, i.e. no potential difference, then there will be no electric field.

9. Earth’s potential is _______
a) Zero
b) Highly positive
c) Highly negative
d) Varies from place to place

Answer: a
Clarification: We consider earth as the storage of infinite positive as well as a negative charge. Therefore, the potential of the earth is always considered to be zero and the potential of every body is measured with respect to earth. That’s why if we connect any charged body to the earth, its potential instantaneously becomes zero.

300+ TOP Ohm’s Law MCQs and Answers | Class12 Physics

Ohm’s Law Multiple Choice Questions

1. Ohm’s law is true for

A. Metallic conductors at low temperature
B. Metallic conductors at high temperature
C. For electrolytes, when current passes through them
D. For diode when current flows

Answer: A. Metallic conductors at low temperature

2. An example of non-ohmic resistance is

A. Diode
B. Tungsten wire
C. Carbon resistance
D. Copper wire

Answer: A. Diode

3. If a current of 5 Amperes flows through the conductor. The number of electrons per second will is

A. 1.6 x 10-19
B. 3.12 x 1019
C. 4 x 1019
D. 7.68 x 1020

Answer: B. 3.12 x 1019

4. In a conductor, if 6-coulomb charge flows for 2 seconds. The value of electric current will be

A. 3 ampere
B. 3 volts
C. 2 amperes
D. 2 volts

Answer: A. 3 amperes

5. An EMF source of 8.0 V is connected to a purely resistive electrical appliance. An electric current of 2.0 A flows through it. What is the resistance offered by the electrical appliances?

A. 4 ohm
B. 6 ohm
C. 2 ohm
D. 3 ohm

Answer: A. 4 ohm

6. A potential difference of 10 V is applied across a conductor whose resistance is 2.5 ohm. What is the value of current flowing through it?

A. 4 amperes
B. 2 amperes
C. 6 amperes
D. 10 amperes

Answer: A. 4 amperes

7. If the conductor resistance is 50 ohm and the current passing through it is 5 A. What is the value of potential difference?

A. 150 V
B. 250 V
C. 50 V
D. 15 V

Answer: B. 250 V

8. When the length of the conductor is doubled and the area of cross-section remains the same then its resistance

A. Remains the same
B. Will be doubled
C. Will become half
D. Will increase by four times

Answer: B. Will be doubled

9. The current passing through a resistor in a circuit is 1 A when the voltage across the same resistor is 10 V. What is the value of current when the voltage across the resistor is 8 V

A. 0.8 A
B. 8 A
C. 80 A
D. 18 A

Answer: A. 0.8 A

10. Two resistors R1 and R2 with resistance 5 ohms and 10 ohms respectively are connected in series. The voltage across R1 is 4 V. What will be the value of current across R2.

A. 0.8 A
B. 8 A
C. 80 A
D. 18 A

Answer: A. 0.8 A

Ohm’s Law objective questions with answers pdf download online exam test

250+ TOP MCQs on Magnetic Field due to a Current Element & Biot-Savart Law | Class12 Physics

Physics Problems for Schools on “Magnetic Field due to a Current Element & Biot-Savart Law”.

1. Give the SI unit of magnetic permeability of free space.
a) T A m-2
b) T A-2 m
c) T A-1 m
d) T A m2
Answer: c
Clarification: Magnetic permeability of free space is a measure of the amount of resistance encountered when forming a magnetic field in a classical vacuum. The SI unit of permeability is weber ampere-1 metre-1 (Wb A-1 m-1) or tesla ampere-1meter (T A-1 m).

2. State the rule that is used to find the direction of field acting at a point near a current-carrying straight conductor.
a) Cork rule
b) The right-hand thumb rule
c) Swimming rule
d) Flemings rule
Answer: b
Clarification: Right-hand thumb rule can be used to find the direction of the magnetic field at a point near a current-carrying conductor. Right hand rule states that, if the thumb of the right hand is in the direction of the current flow then, the curl fingers show the direction of the magnetic field.

3. Give the dimensional formula for magnetic permeability of free space.
a) [M L T-2 A-2]
b) [M2 L T-2 A-2]
c) [M L2 T-2 A-2]
d) [M-1 L T-2 A-2]
Answer: a
Clarification: Magnetic permeability = Magnetic flux density × [Magnetic field strength]-1.
μ = [M L0 T-2 A-1] × [M0 L-1 T0 A1]-1
μ = [M L T-2 A-2].

4. A wire placed along the north-south direction carries a current of 8 A from south to north. Find the magnetic field due to a 1 cm piece of wire at a point 200 cm north-east from the piece.
a) 14 × 10-9 T
b) 1004 × 10-9 T
c) 204.4 × 10-9 T
d) 1.4 × 10-9 T
Answer: d
Clarification: dB = (frac {(mu_0 I dl sin (theta))}{(4pi r^2)}).
dB = (frac {(4 pi , times , 10^{-7} , times , 8 , times , 1 , times , 10^{-2} , times , sin 45^o}{(4 pi , times , 22)})
dB = 1.4 × 10-9 T.

5. Which of the following is not a point of similarity between Biot-Savart law and Coulomb’s law.
a) Both fields depend inversely on the square of the distance from the source to the point of observation
b) They are not a universal law
c) The principle of superposition does not apply to both
d) Both are long-range fields
Answer: c
Clarification: The principle of superposition applies to both fields. This is because the magnetic field is linearly related to its source, namely, the current element and the electrostatic field is related linearly to its source, the electric charge.

6. The magnetic field due to a current element is minimum in a plane passing through the element when it is perpendicular to its axis. State true or false.
a) True
b) False
Answer: b
Clarification: If θ = 90o → sin θ = 1
→ Then dB is maximum.
The magnetic field due to a current element is maximum in a plane passing through the element and perpendicular to its axis.

7. Give the SI unit of the magnetic field from Biot-Savart law.
a) Ampere
b) Tesla
c) Weber
d) Gauss
Answer: b
Clarification: The SI unit of the magnetic field is the tesla (T). One tesla is 107 times the magnetic field produced by a conducting wire of length one meter and carrying a current of one ampere at a distance of one meter from it and perpendicular to it.

Physics Problems for Schools,

250+ TOP MCQs on Electromagnetic Induction – Lenz’s Law and Conservation of Energy | Class12 Physics

Physics Multiple Choice Questions on “Electromagnetic Induction – Lenz’s Law and Conservation of Energy”.

1. A closed-loop move normal to the constant electric field between the plates of a large capacitor. What is the amount of current produced when it is wholly inside the region between the capacitor plates?
a) Maximum
b) Minimum
c) Zero
d) Independent of current
Answer: c
Clarification: No current is produced when it is wholly inside the region between the capacitor plates. Current cannot be induced by changing electric flux. So, the amount of current produced when it is wholly inside the region between the capacitor plates is zero.

2. Which law is used in finding the direction of current in an a.c. generator?
a) Maxwell’s law
b) Lenz’s law
c) Corkscrew law
d) Ampere circuital law
Answer: b
Clarification: In an a.c. generator, induced current due to a change of magnetic flux linked with a closed circuit can be found out using Lenz’s law. Lenz’s law, in electromagnetism, statement that an induced electric current flows in a direction such that the current opposes the change that induced it.

3. Which of the following statement is valid?
a) Lenz’s law is a consequence of the law of conservation of energy
b) Lenz’s law is a consequence of the law of conservation of momentum
c) Lenz’s law is a consequence of the law of conservation of force
d) Lenz’s law is a consequence of the law of conservation of mass
Answer: a
Clarification: Whether a magnet is moved towards or away from a closed coil, the induced current always opposes the motion of the magnet, as predicted by Lenz’s law. Work has to be done in moving the magnet closer to the coil against this force of repulsion. Thus Lenz’s law is valid and is a consequence of the law of conservation of energy.

4. The current in a wire passing normally through the center of a conducting loop is increasing at a constant rate. What is the net current induced in the loop?
a) Indefinite
b) Maximum
c) Minimum
d) Zero
Answer: d
Clarification: The magnetic lines of force due to current are parallel to the plane of the loop. The flux linked with the loop is zero. Hence no current is induced in the loop.

5. Which of the following is found using Lenz’s law?
a) Induced emf
b) Induced current
c) The direction of induced emf
d) The direction of alternating current
Answer: c
Clarification: Lenz’s law is a general law for determining the direction of induced emf and hence that of induced current in a circuit. Lenz’s law states that the direction of an induced emf will be such that if it were to cause a current to flow in a conductor in an external circuit, then that current would generate a field that would oppose the change that created it.

6. Lenz’s law is invalid. State true or false.
a) True
b) False
Answer: b
Clarification: When a magnet is moved towards or away from a closed coil, the induced current always opposes the motion of the magnet, as predicted by Lenz’s law. Thus Lenz’s law is valid and is a consequence of the law of conservation of energy.

7. ‘X’ states that the direction of induced current in a circuit is such that it opposes the cause or the change which produces it. Identify X.
a) Faraday’s law
b) Lenz’s law
c) Maxwell’s law
d) Ampere’s law
Answer: b
Clarification: Lenz’s law states that the direction of induced current in a circuit is such that it opposes the cause or the change which produces it. It is a general law for determining the direction of induced emf and hence that of induced current in a circuit.

8. Identify the law used to find the direction of eddy currents.
a) Lenz’s law
b) Maxwell’s law
c) Ampere’s law
d) Faraday’s law
Answer: a
Clarification: Eddy currents are the currents induced in solid metallic masses when the magnetic flux threading through them changes. Eddy currents also oppose the change in magnetic flux, so their direction is given Lenz’s law.

Physics Multiple Choice Questions and Answers for Class 12,

250+ TOP MCQs on Ray Optics – Refraction | Class12 Physics

Physics Multiple Choice Questions on “Ray Optics – Refraction”.

1. Identify the factor on which the angle of deviation of the prism does not depend.
a) The angle of incidence
b) The material of the prism
c) The angle of reflection
d) The wavelength of light used
Answer: c
Clarification: Factors on which the angle of deviation depends are ➔ the angle of incidence, the material of the prism, the wavelength of light used, and the angle of the prism. So, the factor on which the angle of deviation that does not depend on is the angle of reflection.

2. Calculate the refractive index of the material of an equilateral prism for which the angle of minimum deviation is 60°.
a) (frac {sqrt {3}}{2})
b) √3
c) (frac {1}{2})
d) (frac {1}{sqrt {2}})
Answer: b
Clarification: Refractive index of the prism material is μ = ( frac { { sin frac {(A+delta_m)}{2} } }{ { sin frac {A}{2} } })
μ = ( frac { { sin frac {(60^o+60^o)}{2} } }{ { sin frac {60^o}{2} } } )
μ = √3.

3. Which of the following causes dispersion?
a) Refraction
b) Reflection
c) Total internal reflection
d) Total internal dispersion
Answer: a
Clarification: The reason is that for a given angle f incidence, the reflection is the same for all the wavelengths of white light while the angle of refraction is different for different wavelengths. Therefore, this is how dispersion is caused.

4. What happens to the frequency and the wavelength when light passes from a rarer to a denser medium?
a) Wavelength remains unchanged but frequency changes
b) They are independent
c) Wavelength and frequency changes
d) Wavelength changes but the frequency remain unchanged
Answer: d
Clarification: When light passes from a rarer to a denser medium, the wavelength of light changes but the frequency remains unchanged. Therefore, there is no change in the frequency of light, only the wavelength changes.

5. What is the relative refractive index of water with respect to glass?
a) Unity
b) More than unity
c) Less than unity
d) Zero
Answer: c
Clarification: Absolute refractive index = ( frac {Speed , of , light , in , vacuum}{Speed , of , light , in , the , medium}). The relative refractive index of water with respect to glass is less than unity.

6. The refraction in a water tank makes apparent depth the same throughout.
a) True
b) False
Answer: b
Clarification: No, this statement is false. Apparent depth is maximum for that part of the bottom of the tank which is observed normally. Apparent depth decreases with increasing obliquity. Therefore, the refraction in a water tank does not make apparent depth the same throughout.

7. What will be the color of the sky in the absence of the atmosphere?
a) White
b) Dark
c) Blue
d) Pink
Answer: b
Clarification: The sunlight will not be scattered in the absence of the atmosphere. So the sky will appear dark. So, the sky will no longer be blue in the absence of an atmosphere.

8. On what factor does the normal shift through a refracting medium depend?
a) The thickness of the refracting medium
b) Angle of Prism
c) Angle of deviation
d) Convection
Answer: a
Clarification: The normal shift depends on the thickness of the refracting medium and the refractive index of the material. The normal shift does not depend on the angle of the prism, angle of deviation, and convection.

9. A lens immersed in a transparent liquid is not visible. Under what condition can this happen?
a) Less refractive index
b) Higher refractive index
c) Same refractive index
d) Total internal reflection is zero
Answer: c
Clarification: When the refractive index of the liquid is the same as the lens material, no light will be reflected by the lens and hence it will not be visible. So, the lens immersed in a transparent liquid will not be visible.

10. What is the cause of the blue color of the ocean?
a) Reflection
b) Scattering of light by water molecules
c) Total internal reflection
d) Refraction
Answer: b
Clarification: The blue color of the ocean is due to the preferential scattering of light by water molecules. This is the underlying reason for many other phenomena such as why the sky is blue or the sky is reddish at the time of sunrise or sunset.

250+ TOP MCQs on Particle Nature of Light : The Photon | Class12 Physics

Physics Multiple Choice Questions on “Particle Nature of Light : The Photon”.

1. What is the frequency of a photon whose energy is 66.3 eV?
a) 12.6 × 1016 Hz
b) 91.6 × 1016 Hz
c) 1.6 × 1016 Hz
d) 81.6 × 1016 Hz
Answer: c
Clarification: Frequency can be written as, v = (frac {E}{h})
v = (frac {E}{h})
v = (frac {(66.3 times 1.6 times 10^{-19})}{(6.63 times 10^{-34})} )
v = 1.6 × 1016 Hz.

2. Calculate the energy of a photon of wavelength 6600 angstroms.
a) 0.3 × 10-19 J
b) 3 × 10-19 J
c) 30 × 10-19 J
d) 300 × 10-19 J
Answer: b
Clarification: λ = 6600 angstroms = 6600 × 10-10m.
Energy of photon = (frac {hc}{lambda })
E = (frac {(6.6 times 10^{-34} times 3 times 10^8)}{(6600 times 10^{-10})})
E = 3 × 10-19 J.

3. Among the following four spectral regions, in which of them, the photon has the highest energy in?
a) Infrared
b) Violet
c) Red
d) Blue
Answer: b
Clarification: According to the equation:
E = (frac {hc}{lambda })
Energy is inversely proportional to wavelength. Since a photon in the violet region has the least wavelength, it implies, that photon has the highest energy.

4. What will be the photon energy for a wavelength of 5000 angstroms, if the energy of a photon corresponding to a wavelength of 7000 angstroms is 4.23 × 10-19 J?
a) 0.456 eV
b) 5.879 eV
c) 3.701 eV
d) 1.6 × 10-19 eV
Answer: c
Clarification: (frac {E_2}{E_1} = frac {lambda_1}{lambda_2})
(frac {lambda_1}{lambda_2} = frac {1.4 times 4.23 times 10^{-19}}{1.6 times 10^{-19}}) eV
(frac {lambda_1}{lambda_2}) = 3.701 eV

5. Photons of energy 10.25 eV fall on the surface of the metal emitting photoelectrons of maximum kinetic energy 5.0 eV. What is the stopping voltage required for these electrons?
a) 10 V
b) 4 V
c) 8 V
d) 5 V
Answer: d
Clarification: The required equation is given as:
Stopping voltage = (frac {K_{max}}{e}).
(frac {K_{max}}{e} = frac {5.0 eV}{e})
(frac {K_{max}}{e}) = 5.0 V
Therefore, the stopping potential = 5 V.

6. When a proton is accelerated through 1 V, then its kinetic energy will be 1 V.
a) True
b) False
Answer: b
Clarification: Kinetic energy = qV.
K = qV
K = e × (1V)
K = 1 eV.

7. What is the energy of a photon of wavelength λ?
a) hcλ
b) (frac {hc}{lambda })
c) (frac {lambda }{hc})
d) (frac {lambda h}{c})
Answer: b
Clarification: Energy of a photon, E = hv
V = (frac {c}{lambda }).
Therefore, E = (frac {hc}{lambda })

8. What is the momentum of a photon of wavelength λ?
a) (frac {hv}{c})
b) Zero
c) (frac {hlambda }{c^2})
d) (frac {hlambda }{c})
Answer: a
Clarification: Momentum is given as:
p = mc
p = mc
p = (frac {mc^2}{c})
p = (frac {hv}{c}).

9. Which among the following shows the particle nature of light?
a) Photoelectric effect
b) Interference
c) Refraction
d) Polarization
Answer: a
Clarification: Photoelectric effect can only be explained based on the particle nature of light. This effect is caused due to the ejection of electrons from a metal plate when light falls on it. The others do not show the particle nature of light.

10. What will be the number of photons emitted per second, if the power of the radio transmitter is 15 kW and it operates at a frequency of 700 kHz?
a) 3.24 × 1031
b) 3.87 × 1025
c) 2.77 × 1037
d) 3.24 × 1045
Answer: a
Clarification: Number of photons emitted per second is given as:
N = ( frac {Power}{Energy , of , a , photon} = frac {P}{hv}).
N = ( frac {15 times 10^3}{6.6 times 10^{-34} times 700 times 10^3})
N = 3.24 × 1031