Category: Class 12 Physics Objective Questions

250+ TOP MCQs on Electrostatic Potential due to an Electric Dipole | Class12 Physics

Physics Multiple Choice Questions on “Electrostatic Potential due to an Electric Dipole”.

1. Which one is the correct expression of electric potential on the axial point of a dipole whose dipole moment is p and length is 2l, at a distance of r?
a) (frac {p}{r^2})
b) (frac {p}{l^2})
c) (frac {p}{(r^2-l^2)})
d) (frac {p}{2lr})

Answer: c
Clarification: Electric potentials due to the two point charges of the dipole are (frac {q}{(r-l)}) and (frac {-q}{(r+l)}). Therefore, total potential = (frac {q}{(r-l)} – frac {q}{(r+l)} = frac {2ql}{(r^2-l^2)}). Now 2ql = dipole moment of the dipole=p. Therefore, the expression becomes (frac {p}{(r^2-l^2)}).

2. What is the electric potential at the perpendicular bisector of an electric dipole?
a) Positive
b) Negative
c) Zero
d) Depends on medium

Answer: c
Clarification: Any point on the perpendicular bisector is equidistant from both the charges of the dipole. Therefore, the electric potential at that point is equal and opposite due to the two different charges. Therefore, the total electric potential at that point is zero.

3. The electric potential at an axial point of a dipole is the maximum.
a) True
b) False

Answer: a
Clarification: Electric field at a distance r from a dipole and at an angle θ is p*(frac {cos theta}{r^2}), where p is the dipole moment of that dipole. If θ becomes 0 degrees, cosθ will be the maximum i.e. 1. Therefore, the electric field is the maximum at the axial points.

4. What is the dimension of the dipole moment?
a) [I L T]
b) [I L T-1]
c) [I L2 T]
d) [I T]

Answer: a
Clarification: Dipole moment = charge*length of the dipole. The electric charge has dimensions [I T] and length has dimensions [L]. Therefore, the dipole moment has the dimension [I T L] and has unit C*m of C*cm.

250+ TOP MCQs on Current Electricity – Limitations of Ohm’s Law | Class12 Physics

Physics Multiple Choice Questions on “Current Electricity – Limitations of Ohm’s Law”.

1. Identify the type of conductors whose V-I relationship is linear.
a) Thyristor
b) Non-ohmic conductors
c) Ohmic conductors
d) Superconductors
Answer: c
Clarification: The conductors which obey Ohm’s law are called Ohmic conductors. The linear relationship between voltage and current for these conductors hold good.

2. Which of these relation holds good for an Ohmic conductor?
a) V-I relationship is non-linear
b) The straight-line V-I graph pass through the origin
c) V-I relationship is non-unique
d) V-I relationship depends on the sign of V for the same absolute value of V
Answer: b
Clarification: The conductors which obey Ohm’s law are called Ohmic conductors. The linear relationship between voltage and current for these conductors hold good. The resistance (R=(frac {V}{I})) is independent of the current through the conductor. The magnitude of current changes linearly with voltage. Hence the V-I graph for ohmic conductors is a straight line passing through the origin.

3. Which of the following is an example of an Ohmic conductor?
a) Nichrome
b) Water voltameter
c) Gallium Arsenide
d) Thyristor
Answer: a
Clarification: Most of the metals obey Ohm’s law and they are called ohmic conductors. Whereas semiconductors are non-ohmic. Nichrome metal is an ohmic conductor in which the V-I characteristic has a straight line passing through the origin.

4. Identify the type of conductor represented by the V-I characteristic curve given below.

a) Metallic conductor
b) Water voltameter
c) Thyristor
d) Semiconductor
Answer: d
Clarification: The V-I graph gives the characteristic of a semiconductor. It exhibits non-linear behaviour. After a certain voltage, the current decreases as the voltage increases. Gallium arsenide shows this type of behaviour.

5. Identify the type of conductor represented by the V-I characteristic curve given below.

a) Metallic conductor
b) Thyristor
c) p-n junction diode
d) Semiconductor
Answer: c
Clarification: p-n junction diode consists of p-type and n-type semiconductors. The V-I relationship is non-linear. When a voltage is applied across junction, very little current flows for the fairly high negative voltage and a current begins to flow for much smaller positive (forward) bias.

6. Ohm’s law is a fundamental law of nature. State true or false.
a) True
b) False
Answer: b
Clarification: Ohm’s law is not a fundamental law of nature. Some reasons for the failure of ohm’s law are: Potential difference may vary non-linearly with the current ; The V-I relationship may be non-unique; The variation of current with potential difference may depend upon the sign of the potential difference applied.

7. For small currents, ‘X’ obeys Ohm’s law. But when large currents are passed through the same conductor, it deviates from ohmic behaviour. Identify X.
a) Metallic conductor
b) Gallium arsenide
c) Thyristor
d) p-n junction diode
Answer: a
Clarification: ‘X’ is a metallic conductor. For small currents, it obeys Ohms’s law and its V-I graph is a straight line. But when large currents are passed through the same conductor, it gets heated up and its resistance increases. It no longer obeys Ohm’s law at higher currents.

8. Identify the type of conductor in which the variation of current with potential difference may depend upon the sign of the potential difference applied.
a) Metallic conductor
b) Thyristor
c) Water voltameter
d) p-n junction diode
Answer: d
Clarification: p-n junction diode consists of p-type and n-type semiconductors. The V-I relationship is non-linear. When a voltage is applied across junction, very little current flows for the fairly high negative voltage and a current begins to flow for much smaller positive (forward) bias. The magnitude of variation depends upon the sign of potential difference applied across it.

9. Which of the following is an example of a non-ohmic conductor?
a) Copper
b) Silver
c) Silicon
d) Nichrome
Answer: c
Clarification: Silicon is an example of a non-ohmic conductor. It exhibits non-linear behaviour. After a certain voltage, the current decreases as the voltage increases. The V-I plot for silicon is non-linear.

250+ TOP MCQs on Ampere’s Circuital Law | Class12 Physics

Physics Multiple Choice Questions on “Ampere’s Circuital Law”.

1. Identify the expression for ampere’s circuital law from the following?
a) ∮ B .dl = μ0I
b) ∮ B .dl = ( frac {mu_0}{I})
c) ∮ B .dl = 2μ0I
d) ∮ B .dl = 4μ0I
Answer: a
Clarification: Ampere’s circuital law states that the line integral of magnetic field around any closed path in vacuum is equal to μ0 times the total current passing the closed path. It is given by:
∮ B .dl = μ0I
Ampere’s circuital law is analogous to Gauss’s law in electrostatics.

2. What is the magnetic field inside a pipe?
a) Unity
b) Infinity
c) Zero
d) Two
Answer: c
Clarification: The magnetic field inside a pipe, i.e. inside a hollow cylindrical wire is zero. This is due to the symmetry of the situation (pipe). The pipe can be considered as a series of thin wires arranged in a circle.

3. Which one of the following graphs depict the magnetic field B at a distance r from a long straight wire carrying current varies with distance r?
a)

b)

c)

d)

Answer: b
Clarification: The magnetic field B at a distance r from a long straight wire carrying current I is given by:
B = (frac {mu_0 2I}{4pi r})
B = (frac {mu_0 I}{2pi r})
From this, we understand that the magnetic field is inversely proportional to the distance r.
B ∝ (frac {1}{r})

4. The angle at which a charged particle moves through a magnetic field with a velocity, having zero magnetic force is 180o.
a) True
b) False
Answer: b
Clarification: This statement is false. When a charged particle is moving through a magnetic field with a velocity and has a zero magnetic force, then the angle at which the charged particle should move is 0o or 90o.

5. Which law can ampere’s circuital be derived from?
a) Gauss Law
b) Newton’s Law
c) Kirchhoff’s Law
d) Biot-Savart Law
Answer: d
Clarification: In classical electrodynamics, the magnetic field given by a current loop and the electric field caused by the corresponding dipoles in sheets are very similar, as far as we are far away from the loop, which enables us to deduce Ampere’s magnetic circuital law from the Biot-Savart law easily.

6. A long straight wire of radius x carries a steady current I. The current is uniformly distributed across its cross section. Calculate the ratio of the magnetic field at (frac {x}{4}) and 8x.
a) 4
b) 3
c) 2
d) 1
Answer: c
Clarification: The magnetic field due to a long straight wire of radius x carrying a current I at a point distant r from the axis of the wire is given by:
Bin = (frac {mu_o Ir}{2pi x^2}) (r < x)
Bin = (frac {mu_o I}{2pi r}) (r > x)
The magnetic field at a distance of r = (frac {x}{4}):
B1 = (frac {mu_o I (frac {x}{4})}{2pi x^2} = frac {mu_o I}{8pi x})
The magnetic field at a distance of r = 8x:
B2 = (frac {mu_o I}{2pi (8x)} = frac {mu_o I}{16pi x})
Therefore, (frac {B1}{B2} = frac {frac {mu_o I}{8pi x}}{frac {mu_o I}{16pi x}})
(frac {B1}{B2} = frac {16}{8}) = 2

7. Find the true statement.
a) The force between two parallel current carrying wires is independent of the radii of the wires
b) The force between two parallel current carrying wires is independent of the length of the wires
c) The force between two parallel current carrying wires is independent of the magnitude of currents
d) The force between two parallel current carrying wires is independent of their distance of separation
Answer: a
Clarification: The force between two parallel current carrying wires is independent of the radii of the wires.
From Ampere’s circuital law, the magnitude of the field due to the first conductor can be given by:
B = (frac {mu_o I_1}{2pi d})

The force between the parallel plates is given by:

B= ([frac {mu_o I_1 I_2}{2pi d}] ) L

From this equation, we understand that the force between the parallel current carrying wires does not depend on the ‘radii’.

250+ TOP MCQs on Energy Consideration: A Quantitative Study | Class12 Physics

Physics Online Test for Class 12 on “Energy Consideration: A Quantitative Study”.

1. Identify the significance of energy consideration from the following.
a) Energy consideration provides a link between the force and energy
b) Energy consideration provides a link between the current and energy
c) Energy consideration provides a link between the mechanical and thermal energy
d) Energy consideration provides a link between the voltage and energy
Answer: a
Clarification: Energy consideration provides a link the force and energy. Newton’s problems can also be easily solved with the help of energy consideration. This is the significance of energy consideration.

2. Pick out the expression for power in a rectangular conductor from the following.
a) P=(frac {B^2 l^2 v}{R})
b) P=(frac {B^2 lv^2}{R})
c) P=(frac {B^2 l^2 v^2}{R})
d) P=(frac {B^2 l^2 v^2}{R^2})
Answer: c
Clarification: In a rectangular conductor, the emf (E) is given ➔ Blv
So, current (I) = (frac {E}{R} = frac {Blv}{R})
Power (P) = I2R
P=(frac {B^2 l^2 v^2}{R})
The work done is mechanical and this mechanical energy is dissipated as Joule heat.

3. A circular coil having an area of 3.14 × 10-2 m2 and 30 turns is rotated about its vertical diameter with an angular speed of 70 rad-1 in a uniform horizontal magnetic field of magnitude 5 × 10-2 T. If the coil forms a closed loop of resistance 15 Ω, what is the average power loss due to Joule heating?
a) 0.360
b) 0.362
c) 0.724
d) 0.726
Answer: b
Clarification: Given: N = 30; A = 3.14 × 10-2 m2 ; Angular velocity (ω) = 70 rad-1; B = 5 × 10-2 T; R = 15 Ω
Emf (E) = NABω
E = 30 × 3.14 × 10-2 × 5 × 10-2 × 70
E = 3.297 V
Current (I) = (frac {E}{R})
I = (frac {3.297}{15})
I = 0.2198 A
Average power loss (P) = (frac {I^2 R}{2})
P = 0.21982 × (frac {15}{2})
P = 0.362 W
Therefore, the average power loss due to Joule heating is 0.362 W.

4. Mechanical energy is initially converted into thermal energy in the case of motional emf.
a) True
b) False
Answer: b
Clarification: No, this statement is false. The work done is mechanical and the energy is dissipated as mechanical energy. Then it is initially converted into electric energy and then only finally converted into thermal energy.

5. What is the magnitude of the force on a current carrying conductor of length L in a perpendicular magnetic field B?
a) F=(frac {I}{lB})
b) F=IlB2
c) F=I2 lB
d) F=IlB
Answer: d
Clarification: The current passing through conductor in a perpendicular magnetic field is given as:
I=(frac {varepsilon}{r})
I=(frac {Blv}{r})
Since it is a perpendicular magnetic field, the force ➔ I(l×B) is directed outwards in the direction opposite to the velocity of the rod. The magnitude of the force is given as:
F=IlB
We can also write it as ➔ F=IlB=(frac {B^2 l^2 v}{r})

6. Find the true statement.
a) The force on a current carrying conductor in a perpendicular magnetic field is due to the thermal energy dissipated
b) The force on a current carrying conductor in a perpendicular magnetic field is due to the mechanical energy dissipated
c) The force on a current carrying conductor in a perpendicular magnetic field is due to the current passing through it
d) The force on a current carrying conductor in a perpendicular magnetic field is due to the electrical energy initially converted from mechanical energy
Answer: d
Clarification: The force on a current carrying conductor in a perpendicular magnetic field is due to drift velocity of charges along the current carrying conductor and the consequent Lorentz force acting on it. So, it is due to the electric energy which was the energy that mechanical energy dissipated in the system was initially converted to.

7. Identify the relation between charge flow and change in magnetic flux.
a) ∆Q = ∆ΦB×r
b) ∆Q = 2∆Φ;B r
c) ∆Q = (frac {triangle Phi_B}{r})
d) ∆Q = (frac {triangle Phi_B}{2r})
Answer: c
Clarification: There is a relationship between the charge flow through the circuit and the change in the magnetic flux. According to Faraday’s Law, the induced emf is given by:
|ε|=( frac {triangle Phi_B}{triangle t})
But, |ε|=Ir=(( frac {triangle Q}{triangle t} ) )r
Therefore, the relation is given by:
∆Q = (frac {triangle Phi_B}{r})

Physics Online Test for Class 12,

250+ TOP MCQs on Ray Optics – Refraction at Spherical Surfaces and by Lenses | Class12 Physics

Physics Written Test Questions on “Ray Optics – Refraction at Spherical Surfaces and by Lenses”.

1. How many types of spherical refracting surfaces are there?
a) 2
b) 3
c) 4
d) 5
Answer: a
Clarification: The portion of a refracting medium, whose curved surface forms the part of a sphere, is known as a spherical refracting surface. Spherical refracting surfaces are of two types, namely, convex refracting spherical surface and concave refracting spherical surface.

2. Which among the following is a portion of a transparent refracting medium bound by one spherical surface and the other plane surface?
a) Concave mirror
b) Plane mirror
c) Lens
d)Prism
Answer: c
Clarification: Lens is a portion of a transparent refracting medium bound by two spherical surfaces or one spherical surface and the other plane surface. Lenses are divided into two classes, namely, convex lens or converging lens, and concave lens or diverging lens.

3. X is thicker in the middle than at the edges, whereas, Y is thicker at the edges than in the middle. Identify ‘X’ and ‘Y’.
a) X = concave lens; Y = convex lens
b) X = convex lens; Y = concave lens
c) X = plane lens; Y = convex lens
d) X = concave lens; Y = plane lens
Answer: b
Clarification: When the lens is thicker in the middle than at the edges, then it is convex lens. When the lens is thicker at the edges than in the middle, then it is concave lens. Therefore, X = convex lens, and Y = concave lens.

4. Sign conventions for spherical refracting surface are the same as those for spherical mirrors.
a) True
b) False
Answer: a
Clarification: Yes, this is a true statement. Sign conventions for spherical refracting surface are the same as those for spherical mirrors. The only difference is that instead of the pole of the mirror, we use the optical center of a lens.

5. Identify the Lens Maker’s Formula.
a) f=(μ-1)( ( frac {1}{R_1} – frac {1}{R_2} ) )
b) (frac {1}{f})=(μ-1)( ( frac {1}{R_1} – frac {1}{R_2} ) )
c) (frac {1}{f})=(μ+1)( ( frac {1}{R_1} – frac {1}{R_2} ) )
d) (frac {1}{f})=(μ-1)( ( frac {1}{R_1} + frac {1}{R_2} ) )
Answer: b
Clarification: The lens maker formula may be defined as the formula which gives the relationship for calculating the focal length in terms of the radii of the two curvatures of the lens. It is given as:
(frac {1}{f})=(μ-1)( ( frac {1}{R_1} – frac {1}{R_2} ) )
Where R1 and R2 are radii of curvature of the two surfaces of the lens and μ is the refractive index of the material of the lens w.r.t. medium in which the lens is placed.

6. According to the thin lens formula, which one of the following is true regarding the focal length of the lens?
a) f is positive for concave lens
b) f is negative for convex lens
c) f is positive for a diverging lens
d) f is negative for concave lens
Answer: d
Clarification: Thin lens formula is given as:
(frac {1}{f}=frac {1}{v}-frac {1}{u})
Where u = distance of the object from the optical center of the lens; v = distance of the image from the optical center of the lens; f = focal length of a lens. According to the thin lens formula, f is positive for converging or convex lens and f is negative for a diverging or concave lens.

7. What is the SI unit of power of a lens?
a) Watts
b) Unit less
c) Diopter
d) Joule
Answer: c
Clarification: The power of a lens is defined as the reciprocal of its focal length in meters. The expression is given as:
P = (frac {1}{f})
Where P is the power and f is the focal length in meters. Lens surface power can be found with the index of refraction and the radius of curvature. The SI unit of power of a lens is diopter.

8. The power of a convex lens is negative.
a) True
b) False
Answer: b
Clarification: No, this is a false statement. The power of a convex lens is positive as a convex lens has a positive focal length, while the power of a concave lens is negative as a concave lens has a negative focal length.

9. Calculate the focal length of a biconvex lens if the radii of its surfaces are 50 cm and 20 cm, and index of refraction of the lens glass = 1.2.
a) 0.014 cm
b) 0.715 cm
c) 0.14 cm
d) 71.5 cm
Answer: d
Clarification: Given: R1 = 50 cm; R2 = 20 cm; μ2 = 1.2
Required equation ➔ (frac {1}{f})=(μ-1)( ( frac {1}{R_1} – frac {1}{R_2} ) )
(frac {1}{f}=( frac {mu 2}{mu 1} -1 ) ( frac {1}{R_1} – frac {1}{R_2} ) )
(frac {1}{f}=( frac {1.2}{1} -1 ) ( frac {1}{50} – frac {1}{-20} ) )
(frac {1}{f})=(0.2)( ( frac {1}{50} + frac {1}{20} ) )
(frac {1}{f})=(0.2)( ( frac {20}{1000} + frac {50}{1000} ) )
(frac {1}{f})=(0.2)( ( frac {7}{100} ) )
(frac {1}{f})=0.014
f=+71.5 cm

10. A lens has a focal length of 10 cm. Where the object should be placed if the image is to be 40 cm in the positive direction from the lens?
a) 12 cm
b) 40 cm
c) 13 cm
d) 0.075
Answer: c
Clarification: Given: f = + 20 cm; v = + 40 cm
The required equation ➔
(frac {1}{f}=frac {1}{v}-frac {1}{u})
(frac {1}{10}=frac {1}{10}-frac {1}{u})
(frac {1}{u}=frac {1}{10}-frac {1}{40} = frac {40-10}{400})
(frac {1}{u}=frac {30}{400}=frac {3}{40})
u=(frac {40}{3})=13.3 cm

11. Find the magnification of the lens if the focal length of the lens is 10 cm and the size of the image is -30 cm.
a) 2
b) 3
c) 4
d) 5
Answer: a
Clarification: Given: f = +10 cm; v = -30 cm
Required equations ➔
(frac {1}{f}=frac {1}{v}-frac {1}{u})
m=(frac {size , of , the , image}{size , of , the , object}=frac {v}{u})
(frac {1}{u}=frac {1}{v}-frac {1}{f}=frac {1}{-30}-frac {1}{10} = frac {-(10+30)}{300} = frac {-40}{300}= frac {-2}{15})
u=(frac {-15}{2}) = -7.5 cm
m=(frac {v}{u}= – frac {30}{-7.5}=frac {-300}{-75})= +4
Therefore, the object should be placed at a distance of 7.5 cm from the lens to get the image at a distance of 30 cm from the lens. It is four times enlarged and is erect.

12. A lens has a power of +3 diopters in air. What will be the power of the lens if it is completely immersed in water? Given, μg = (frac {3}{2}) and μw = (frac {4}{3}).
a) 5 D
b) 3 D
c) 1 D
d) (frac {3}{4}) D
Answer: c
Clarification: Pw=(frac {mu_w}{f_w})=μw( [ frac {mu_g}{mu_w} -1 ] [ frac {1}{R1} – frac {1}{R2} ])……………………………..1
Pa=(frac {1}{f_a})=[μg-1]([ frac {1}{R1} – frac {1}{R2} ])……………………………………2
Dividing 1 and 2
We get ➔ (frac {P_w}{P_a} = frac {mu_g-mu_w}{μ_g-1})
Pw=(frac {frac {3}{2}-frac {4}{3}}{frac {3}{2}-1}) × 3
Pw=(frac {frac {1}{6}}{frac {1}{2}}) × 3
Pw=(frac {2times 3}{6})=1 D
Therefore, the power of the lens gets altered inside water.

Physics Written Test Questions and Answers for Class 12,

250+ TOP MCQs on Davisson and Germer Experiment | Class12 Physics

Physics Multiple Choice Questions on “Davisson and Germer Experiment”.

1. Which theory is confirmed by the Davisson – Germer experiment?
a) de – Broglie theory
b) Newton’s theory
c) Einstein’s theory
d) Planck’s theory
Answer: a
Clarification: Davisson and Germer experiment proves the concept of wave nature of matter particles. The Davisson–Germer experiment provides a critically important confirmation of the de-Broglie hypothesis, which said that particles, such as electrons, are of dual nature.

2. Which of the following is used in the Davisson – Germer experiment?
a) Double slit
b) Single slit
c) Electron gun
d) Electron microscope
Answer: c
Clarification: Davisson – Germer experiment uses an electron gun to produce a fine beam of electrons which can be accelerated to any desired velocity by applying a suitable voltage across the gun. The others mentioned do not find an application here.

3. Which crystal is used in the Davisson – Germer experiment?
a) Aluminum
b) Nickel
c) Cobalt
d) Zinc
Answer: b
Clarification: The crystal used in the Davisson – Germer experiment is nickel. A fine beam of electrons is made to fall on the surface of the nickel crystal. As a result, the electrons are scattered in all directions by the atoms of the crystal.

4. Intensity is different for different angles of scattering in the Davisson – Germer experiment.
a) True
b) False
Answer: a
Clarification: Yes, this statement is true. When the graphs are drawn showing the variation of intensity of the scattered electrons with the angles of scattering at different accelerating voltages, it is found that the intensity is different for different angles of scattering.

5. Identify the expression for Bragg’s law from the following.
a) 2d cos⁡θ=nλ
b) 2d sin⁡θ=nλ
c) 2d sinθ=(frac {n}{lambda })
d) 2d cosθ⁡=(frac {n}{lambda })
Answer: b
Clarification: Bragg’s law is a special case of Laue diffraction, it gives the angles for coherent and incoherent scattering from a crystal lattice. The expression for Bragg’s law is given as:
2d sin⁡θ=nλ