Category: Class 12 Physics Objective Questions

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,

Physics Assessment Questions for Schools on “Current Electricity – Drift of Electrons and the Origin of Resistivity”.

1. What is the SI unit of mobility?
a) Vm^-1
b) m²V^-1s^-1
c) mV^-2
d) m²V^-2s^-1
Answer: b
Clarification: The SI unit of mobility is m²V^-1s^-1.
Mobility = (frac {Drift , velocity}{Electric , field})
SI unit = (frac {ms^{-1}}{Vm^{-1}}) = m²V^-1s^-1.

2. Consider a conductor of length 0.5 m. A potential difference of 20V is applied across this conductor. If the drift velocity of electrons is given as 5.0 × 10^-4ms^-1, then determine the mobility of the electrons.
a) 5.25 × 10¹⁸m²V^-1s^-1
b) 5.25 × 10^-18m²V^-1s^-1
c) 1.25 × 10^-5 m²V^-1s^-1
d) 1.25 × 10⁶m²V^-1s^-1
Answer: c
Clarification: Mobility = (frac {Drift , velocity}{Electric , field}).
Electric field = (frac {Voltage}{length} = frac {20}{0.5}) = 40Vm^-1.
Mobility = (frac {Drift , velocity}{Electric , field} = frac {5.0 times 10^{-4}}{40}) = 1.25 × 10^-5 m²V^-1s^-1.

3. Calculate the drift velocity of free electrons if a current of 5 A is maintained in a conductor of cross-section 10^-2m². The number density of free electrons is 5 × 10²⁰m^-3.
a) 6.25 ms^-1
b) 5.25 ms^-1
c) 2.25 ms^-1
d) 12.25 ms^-1
Answer: a
Clarification: Drift velocity = (frac {Current}{(number , density , of , electrons , times , charge , on , electron , times , Area)}).
(frac {5}{(5 times 10^{20} times 1.6 times 10^{-19} times 10^{-2})})
= 6.25 ms^-1.

4. Predict the effect of temperature of the conductor on the drift velocity of electrons.
a) Drift velocity varies linearly with temperature
b) Drift velocity does not depend on the temperature
c) Drift velocity increases with increasing temperature
d) Drift velocity decreases with increasing temperature
Answer: d
Clarification: On increasing the temperature of a conductor, the value of resistivity of its material increases. Resistivity is indirectly proportional to drift velocity. Therefore, the drift velocity of electrons decreases with the increasing temperature of the conductor.

5. A potential difference of 100 V is applied across a conductor of length 50cm. Calculate the drift velocity of electrons if the electron mobility is 9 × 10^-5 m²V^-1s^-1.
a) 0.001 ms^-1
b) 1.800 ms^-1
c) 0.018 ms^-1
d) 0.180 ms^-1
Answer: c
Clarification: Drift velocity = Mobility × Electric field.
It can be rearranged as,
Drift velocity = (frac {(mobility , times , Potential , difference)}{length}).
Drift velocity = (frac {(9 × 10^{-5} times 100)}{0.5}) = 0.018 ms^-1.

6. The mobility of a charge carrier is the drift velocity acquired by it in a unit cross-sectional area.
a) True
b) False
Answer: b
Clarification: The mobility of a charge carrier is the drift velocity acquired by it in a unit electric field.
Mobility=(frac {Drift , velocity}{Electric , field}). SI unit = (frac {ms^{-1}}{Vm^{-1}}) = m²V^-1s^-1.

7. Predict the effect of length of conductor on drift velocity of electrons.
a) Drift velocity varies linearly with the length of conductor
b) Drift velocity does not depend on the length of conductor
c) Drift velocity increases with the increasing length of conductor
d) Drift velocity decreases with the increasing length of conductor
Answer: d
Clarification: The drift velocity of electrons decreases when the length of the conductor is increased.
Drift velocity = (frac {Potential , difference}{(number , of , electrons , times , charge , of , electron , times , length , of , conductor , times , density)}).

8. Calculate the drift velocity of free electrons if a current of 200 A is maintained in a conductor of cross-section 10^-3m². The number density of free electrons is 96 × 10²⁰ m^-3.
a) 130.20 ms^-1
b) 13.020 ms^-1
c) 1.3020 ms^-1
d) 0.1302 ms^-1
Answer: a
Clarification: Drift velocity = (frac {Current}{(number , density , of , electrons , times , charge , on , electron , times , Area)}).
=(frac {200}{(96 times 10^{20} times 1.6 times 10^{-19} times 10^{-3})})
= 130.20 ms^-1.

Physics Assessment Questions for Schools,

Physics Written Test Questions for Schools on “Magnetic Field on the Axis of a Circular Current Loop”.

1. Pick out the expression for magnetic field strength at any point at the center of a circular loop from the following?
a) B = ( [ frac {mu_o I}{4 pi r} ] int_0^{2pi r} )dl sin⁡90
b) B = ( [ frac {mu_o I}{4 pi r} ] int_0^{2pi r} )dl sin⁡45
c) B = ( [ frac {mu_o I}{4 pi r} ] int_0^{2pi r} )dl sin⁡30
d) B = [μ_o × 4πr]( int_0^{2pi r} )dl sin⁡90
Answer: a
Clarification: The magnetic field strength at any point at center of circular loop carrying current I and radius r is given as:
B = ( [ frac {mu_o I}{4 pi r} ] int_0^{2pi r} )dl sin⁡90
It can also be expressed as → B = ( [ frac {mu_o I}{4 pi r} ] ) × 2πr or B = ( [ frac {mu_o I}{2 r} ] ) direction.
It is inwards if the current is flowing in the clockwise direction and it is onwards if the current is flowing in the anticlockwise direction.

2. Identify the expression for the magnetic field on the axis of circular loop.
a) B=( frac {mu_o Ir}{4 pi big ( (r^2 , + , x^2)^{frac {1}{2}} big ] } ) × 2πr
b) B=( frac {mu_o Ir}{4 pi big ( (r^2 , + , x^2)^{frac {3}{2}} big ] } )
c) B=( frac {mu_o Ir}{4 pi big ( (r^2 , + , x^2)^{frac {3}{2}} big ] } ) × 2πr
d) B=( frac {mu_o Ir}{4 pi big ( (r^2 , + , x^2)^{frac {1}{2}} big ] } ) × 4πr
Answer: c
Clarification: The expression for magnetic field on the axis of circular loop is given as:
B=( frac {mu_o Ir}{4 pi big ( (r^2 , + , x^2)^{frac {3}{2}} big ] } ) × 2πr
It is towards the loop if current in it is in clockwise direction and it is away from the loop if current in it is in anticlockwise direction.

3. The magnetic field due to a current carrying circular loop of radius 4 cm at a point on the axis at a distance of 7 cm from the center is 48 μT. What will be the value at the center of the loop?
a) 390 μT
b) 393 μT
c) 395 μT
d) 397 μT
Answer: b
Clarification: Field along axis of coil → B =( frac {mu_o IR^2}{2 big ( (R^2 , + , x^2)^{frac {3}{2}} big ] } )
At the coil of the coil → B₁ = ( frac {mu_o I}{2R})
(frac {B_1}{B} = frac {mu_o I}{2R} , times , frac {2 big ( (R^2 , + , x^2)^{frac {3}{2}} big ] }{mu_o IR^2} = frac {2 big ( (R^2 , + , x^2)^{frac {3}{2}} big ] }{R^3})
B1 = (frac {48 [4^2 + 7^2]^{frac {3}{2}}}{4^3})
B1 = (frac {48 times (65)^{frac {3}{2}}}{4^3})
B1 = 131.01 × 3
B1 = 393.03 = 393 μT

4. When an arc of a circle of radius R subtends an angle of (frac {pi }{4}) at the center, and carries a current I, the magnetic field at the center is B = (frac {mu_o I}{16R}).
a) True
b) False
Answer: a
Clarification: Magnetic field at the center of a circular arc of radius R, carries a current I and makes an angle θ at the center is given by
B = (frac {mu_o I theta}{4 pi R})
In this case θ = (frac {pi }{4})
B = (frac {mu_o I (frac {pi }{4})}{4 pi R})
B = (frac {mu_o I}{16 R})

5. A horizontal overhead power line carries a current of 100A in east to west direction. What is the magnitude and direction of the magnetic field due to the current, 2 m below the line?
a) 3 x 10^-5 T
b) 2 x 10^-5 T
c) 1 x 10^-5 T
d) 4 x 10^-5 T
Answer: c
Clarification: Given: I = 100A; r = 2 m; (frac {mu_o}{4 pi }) = 10^-7
The required equation → B = ((frac {mu_o}{4 pi }) , times , frac {2I}{r})
B = 10^-7 × 2 × (frac {100}{2})
B = 10^-7 × 100
B = 1 × 10^-5 T
Therefore, the magnitude of the magnetic field is 1 × 10^-5 T and the direction of the magnetic field is south.

Physics Written Test Questions for Schools,