Category: Class 12 Physics Objective Questions

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 = 90^o → 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,

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,

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.

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

Physics MCQs for IIT JEE Exam on “Semiconductor Electronics – Application of Junction Diode as a Rectifier”.

1. What is a rectifier used for?
a) Convert ac voltage to dc voltage
b) Convert dc voltage to ac voltage
c) Measure resistance
d) Measure current
Answer: a
Clarification: A rectifier is based on the fact that a forward bias p-n junction conducts and a reverse bias p-n junction does not conduct electricity. The rectifier is used to convert alternating current voltage (ac) to direct current voltage (dc).

2. How many main types of rectifiers are there?
a) 1
b) 5
c) 2
d) 4
Answer: c
Clarification: Rectifier is a device that does the process of rectification. This means that rectifiers straighten the direction of the current flowing through it. There are mainly 2 types of rectifiers, namely, full-wave rectifiers and half-wave rectifiers.

3. What is the ripple factor for a half-wave rectifier?
a) 2.0
b) 1.21
c) 0.482
d) 0.877
Answer: b
Clarification: For a half-wave rectifier,
I_rms=(frac {I_m}{2}); I_dc=(frac {I_m}{pi })
r = (sqrt {frac {(frac {I_m}{2} )^2}{(frac {I_m}{pi } )^2} – 1})
r=1.21

4. The ripple frequency of a full-wave rectifier is twice to that of a half-wave rectifier.
a) True
b) False
Answer: a
Clarification: Yes, this is a true statement. The ripple frequency is doubled in a full-wave rectifier because we have to rectify both the positive and negative sides of the waveform. For example, if the input frequency is 50 Hz, then the ripple frequency of a full-wave rectifier is 100 Hz.

5. Identify the expression for rectification efficiency from the following.
a) η=(frac {ac , input , power , from , transformer , secondary}{dc , power , delivered , to , load})
b) η=dc power delivered to load × ac input power from transformer secondary
c) η=dc power delivered to load + ac input power from transformer secondary
d) η=(frac {dc , power , delivered , to , load}{ac , input , power , from , transformer , secondary})
Answer: d
Clarification: The rectification efficiency tells us what percentage of the total input ac power can be converted into useful dc output power. The expression for rectification efficiency is given as:
η=(frac {dc , power , delivered , to , load}{ac , input , power , from , transformer , secondary})

6. What is the form factor for a full-wave rectifier?
a) 1.11
b) 1.57
c) 2.62
d) 0.453
Answer: a
Clarification: For a full-wave rectifier,
I_rms=(frac {I_m}{sqrt {2}}); I_dc=(frac {2I_m}{pi})
Form factor=(frac {I_{rms}}{I_{dc}})
Form factor=(frac {frac {I_m}{sqrt {2}}} {frac {2I_m}{pi}})
Form factor=(frac {pi }{2sqrt {2}})=1.11

7. An alternating voltage of 360 V, 50 Hz is applied to a full-wave rectifier. The internal resistance of each diode is 100 W. If R_L = 5 kW, then what is the peak value of output current?
a) 0.9 A
b) 0.07 A
c) 0.097 A
d) 1.097 A
Answer: c
Clarification: The required equation is as follows:
I_peak=I_rms × √2=(frac {V_{rms} times sqrt {2}}{R_L+2r_p})
I_peak=(frac {360 times sqrt {2}}{5000 + 200})
I_peak=(frac {360 times 1.414}{5200})
I_peak=0.097 A

8. Find the value of output direct current if the peak value of output current is given as 0.095 A.
a) 0.6
b) 0.060
c) 0.05
d) 6.06
Answer: b
Clarification: Given: I₀ = 0.095 A
The required equation is ➔ I_DC = (frac {(2 times I_O)}{pi })
I_DC = (frac {(2 times 0.095)}{3.14})
I_DC = 0.060 A.

9. What is the rms value of output current if the peak value of output current is given as 0.092 A?
a) 0.65 A
b) 6.5 A
c) 0.45 A
d) 0.065 A
Answer: d
Clarification: Given: I₀ = 0.092 A
The required equation is ➔ I_rms = (frac {I_O}{sqrt {2}})
I_rms = (frac {0.092}{1.414})
I_rms = 0.065 A

10. Calculate the value of peak reverse voltage (P.I.V.) if the full-wave rectifier has an alternating voltage of 300 V.
a) 849 V
b) 800 V
c) 750 V
d) 870 V
Answer: a
Clarification: Given: E_rms = 300 V
The required equation is ➔ P.I.V. = 2 × E₀ or P.I.V. = 2√2 × E_rms
P.I.V. = 2√2 × 300 V
P.I.V. = 848.52 V ≈ 849 V.

Physics MCQs for IIT JEE Exam,