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WBCHSE Class 12 Physics MCQs

Atom Multiple Choice Questions

Question 1. The binding energy of an H atom, considering an electron moving around a fixed nucleus (proton), is B = \(\frac{m e^4}{8 n^2 \epsilon_0^2 h^2}\) ( m = mass of an electron)  If one decides to work in a frame of reference where the electron is at rest, the proton would be moving around it. By similar arguments, the binding energy would be  B = \(\frac{m e^4}{8 n^2 \epsilon_0^2 h^2}\) ( m = mass of proton). This last expression is not correct because

1. n would not be integral
2. Bohr’s quantization applies only to electrons
3. The frame in which the electron is at rest Is not inertial
4. The motion of the proton would not be in circular orbits, even if approximately

Answer: 3. The frame in which the electron is at rest Is not inertial

Question 2. The simple Bohr model cannot be directly applied to calculate the energy levels of an atom with many electrons. This is because

1. Of the electrons not being subject to a central force
2. Of the electrons colliding with each other
3. Of screening effects
4. The force between the nucleus and an electron will no longer be given by Coulomb’s law

Answer:  1. Of the electrons not being subject to a central force

Question 3. O₂ molecules consist of two oxygen atoms. In the molecule, nuclear force between the nuclei of the two atoms

1. Is not important because nuclear forces are short-ranged
2. Is as important as electrostatic force for binding the two atoms
3. Cancels the repulsive electrostatic force between the nuclei
4. Is not important because oxygen nuclei have an equal number of neutrons and protons

Answer: 1. Is not important because nuclear forces are short-ranged

Read and Learn More Class 12 Physics Multiple Choice Questions

Question 4. For the ground state, the electron in the H atom has an angular momentum =ti, according to the simple Bohr model. Angular momentum is a vector and hence there will be infinitely many orbits with the vector pointing in all possible directions. It is not true,

1. Because the Bohr model gives incorrect values of angular momentum
2. Because only one of these would have a minimum energy
3. Angular momentum must be in the direction of the spin of electrons
4. Because electrons go around only in horizontal orbits

Answer: 1. Because the Bohr model gives incorrect values of angular momentum

Question 5. Two H atoms in the ground state collide inelastically. The maximum amount by which their combined kinetic energy is reduced is

1. 10.20 eV
2. 20.40 eV
3. 13.6 eV
4. 27.2 eV

Answer:  1. 10.20

WBBSE Class 12 Atom MCQs

Question 6. An ionized H molecule consists of an electron and two protons. The protons are separated by a small distance of the order of angstrom. In the ground state

1. The electron would not move in circular orbits
2. The energy would be 24 times that of a H -atom
3. The electrons, orbits would go around the protons
4. The molecule soon decays in a proton and a neutron

Answer: 1 and 3

Question 7. Consider aiming a beam of free electrons towards free protons. When they scatter, an electron, and a proton cannot combine to produce a H atom

1. Because of energy conservation
2. Without simultaneously releasing energy in the form of radiation
3. Because of momentum conservation
4. Because of angular momentum conservation

Answer: 1 and 2

Question 8. The Bohr model for the spectra of a H atom

1. This will not apply to hydrogen in the molecular form
2. Will not be applicable as it is for a He atom
3. Is valid only at room temperature
4. Predicts continuous as well as discrete spectral lines

Answer: 1 and 2

WBCHSE class 12 physics MCQs

Question 9. The simple Bohr model does not apply to the He atom because

1. He⁴ is an inert gas
2. He⁴ has neutrons in the nucleus
3. He⁴ has one more electron
4. Electrons are not subject to central forces

Answer: 3 and 4

Question 10. The Balmer series for the H atom can be observed

1. If we measure the frequencies of light emitted when an excited atom falls to the ground state
2. If we measure the frequencies of light emitted due to transitions between excited states and the first excited state
3. In any transition in an H atom
4. As a sequence of frequencies with the higher frequencies getting closely packed

Answer: 2 And 4

Question 11. Let \(E_n=-\frac{1}{8 \epsilon_0^2} \frac{m e^4}{n^2 h^2}\) be the energy of the nth level of the H atom If all the H atoms are in the ground state and radiation of frequency (E₂ – E₁)/h falls on it

1. It will not be absorbed at all
2. Some atoms will move to the first excited state
3. All atoms will be excited to the n = 2 state
4. No atoms will make a transition to the n = 3 state

Answer: 2 And 4

Question 12. According to Rutherford’s atomic model, which of the following is correct?

1. Atom is stable
2. The majority of space in an atom is empty
3. E = hf
4. None of these

Answer: 2. The majority of space in an atom is empty

Question 13. Which of the following was used in Rutherford’s experiment

1. Aluminium
2. Platinium foil
3. Silver foil
4. Gold foil

Answer: 4. Gold foil

Common MCQs on Atomic Models

Question 14. What is the velocity of the electron in the first Bohr orbit of a hydrogen atom?

1. 3 × 10⁸ m.s^-1
2. 2.19 × 10⁶ m. s^-1
3. 3 × 10⁷ m. s^-1
4. 3 × 10⁷ m. s^-1

Answer:  2. 2.19 × 10⁶ m. s^-1

Question 15.  How many times will the electron in the first Bohr orbit of a hydrogen atom revolve in 1 s?

1. 6.58 × 10¹⁵
2. 4. 13 × 10¹⁶
3. 1.64 × 10¹⁵
4. 4.13× 10¹⁵

Answer: 1. 6.58 × 10¹⁵

Question 16. The total energy of an electron in the ground state of a hydrogen atom is

1. Zero
2. 13.6 eV
3. – 13.6 eV
4. -13.6 J

Answer: 3. – 13.6 eV

Question 17. What is the minimum energy (in eV) necessary to liberate an electron from the ground state of a Li⁺⁺ ion (Z = 3), according to Bohr’s theory?

1. 1.51
2. 13.6
3. 40.8
4. 122.4

Answer: 4. 122.4

Question 18. An electron in a hydrogen atom undergoes a transition from n₁ to n₂, where n₁ and n₂ are the principal quantum numbers of two given states. According to Bohr’s theory, if the period ofthe electron in the initial state is eight times that in the final state, the possible values of n₁ and n₂ will be, respectively

1. 4 and 2
2. 8 and 2
3. 8 and 1
4. 6 and 4

Answer: 4 and 2

WBCHSE class 12 physics MCQs

Question 19. The atomic number of Helium is 2. What is the energy of the ground state of Helium ion having a single positive charge?

1. -13.6 eV
2. -54.4 eV
3. -40.8 eV
4. -122.4 eV

Answer: 2. -54.4 eV

Question 20. The ionization potential of an atom is 24.6 V. How much energy is required to ionize it?

1. 24.6 eV
2. 2.46 eV
3. 246 eV
4. 0.246 eV

Answer: 1. 24.6 eV

Question 21. In Bohr’s model of the hydrogen atom

1. The radius of the n-th orbit is directly proportional to n²
2. The total energy of an electron in the n-th orbit is inversely proportional to n
3. The angular momentum of an electron in any orbit is an integral multiple of h
4. The potential energy of an electron in any orbit is more than its kinetic energy

Answer: 1. The radius of the n-th orbit is directly proportional to n²

Question 22. The ground state energy of the hydrogen atom is -13.6 eV. If the electron in this atom jumps from the fourth level to the second level, what will be the wavelength of the emitted radiation?

1. 2918 Å
2. 1824 Å
3. 4863 Å
4. 3824 Å

Answer: 3. 4863 Å

Question 23. The electron in a hydrogen atom has been excited to the nth state. What is the maximum number of spectral lines that may be emitted by the atom when the electron transits to the ground state?

1. \(\frac{1}{6} n(n-1)(n-2)\)
2. n
3. n(n-1)
4. \(\frac{1}{2} n(n-1)\)

Answer: 4. \(\frac{1}{2} n(n-1)\)

Question 24. If the electron in a hydrogen atom is raised to the third orbit, how many photons of different energies may be emitted?

1. 1
2. 2
3. 3
4. 4

Answer: 3. 3

Question 25. According to Bohr’s model, the ratio of the energies of the electron in the first orbit of the hydrogen atom and the He⁺ atom is

1. 1:2
2. 4:1
3. 1:4
4. 1:9

Answer: 3. 1:4

Dual nature of matter and radiation class 12 MCQs 

Question 26. The quantized physical quantity of an atomic electron, according to Bohr’s model, is

1. Linear velocity
2. Angular velocity
3. Linear momentum
4. Angular momentum

Answer: 4. Angular momentum

Practice MCQs on Atomic Theory

Question 27. A one-electron atom has an energy of -3,4eV. The kinetic energy of the electron is

1. -3.4 eV
2. 3.4 eV
3. -6.8 eV
4. 6.8 eV %

Answer: 2. 3.4 eV

Question 28. A one-electron atom has an energy of -3.4 eV. The potential energy of the electron is

1. -3.4 eV
2. 3.4 eV
3. -6.8 eV
4. 6.8 eV

Answer: 3. -6.8 eV

Question 29. An electron is in the third orbit of a hydrogen atom. If h = 6.6 × 10^-34J.s, its orbital angular momentum is

1. 1.98 × 10³³J.s
2. 2.2 × 10^-34J.s
3. 3.15 × 10^-34J.s
4. 1.05 × 10^-34J.s

Answer: 3. 3.15 × 10^-34J.s

Question 30. The ratio between the radii of the fourth and the second electron orbits of a hydrogen atom is

1. 2:1
2. 4:1
3. 8:1
4. 16:1

Answer: 2. 4:1

Question 31. The ratio between the electron velocities in the second and the third orbits of a hydrogen atom is

1. 4:9
2. 2:3
3. 3:2
4. 9:4

Answer: 3. 3:2

Question 32. The ratio of the electron revolution frequencies in the first and in second orbits of a hydrogen atom is

1. 8:1
2. 4:1
3. 2: 1
4. 1:4

Answer: 8:1

Dual nature of matter and radiation class 12 MCQs 

Question 33. In an inelastic collision, an electron excites as a hydrogen atom from its ground state to a M -shell state. A second electron collides instantaneously with the excited hydrogen atom in the M -state and ionizes it. At least how much energy does the second electron transfer to the atom in the M – state?

1. + 3.4 eV
2. +1.51 eV
3. -3.4 eV P
4. -1.51 eV

Answer: 2. +1.51 eV

Hint: Total energy of the electrons in n -th state hydrogen of the atom,

⇒ \(E_n=-\frac{13.6}{n^2} \mathrm{eV}\)

For M state

E_m = \(-\frac{13.6}{3^2}\)

= 0 –  (-1.51)= -1.51 eV

∴ The least energy transferred to the atom

= -1.51 eV

Class 12 physics dual nature questions 

Question 34. The energy required for the electron excitation In Li²⁺ from the first to the thin! Bohr orbit Is

1. 36.3 eV
2. 108.8 eV
3. 122.4 eV
4. 12.1 eV

Answer: 2. 108.8 eV

Hint: Total energy of the nth state of an atom

⇒ \(-13.6 \frac{Z^2}{n^2} \mathrm{eV}\)

For Li²⁺ Z = 3

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Question 35. The wavelength of the first line of the Lyman series for a hydrogen atom is equal to that of the second line of the Balmer series for a hydrogen-like ion. The atomic number Z of hydrogen-like ion is

1. 3
2. 4
3. 1
4. 2

Answer: 4. 2

Hint:  The wavelength of the first line of the Lyman series of hydrogen atoms is given by

⇒ \(\frac{1}{\lambda}=R\left(\frac{1}{1^2}-\frac{1}{2^2}\right)\)

Or, \(\lambda=\frac{4}{3 R}\)

The wavelength of the second line of the Balmer series for hydrogen-like atoms is given by

⇒ \(\lambda=\frac{4}{3 R}\)

According to the question

⇒  \(\frac{1}{\lambda^{\prime}}=Z^2 R\left(\frac{1}{2^2}-\frac{1}{4^2}\right)\)

⇒  \(\lambda^{\prime}=\frac{16}{3 Z^2 R}\)

Or Z =2

According to the question

Or = \(\frac{4}{3 R}=\frac{16}{3 Z^2 R}\)

Or, Z = 2

Important Formulas in Atomic Physics

Question 36. An electron in the hydrogen atom jumps from the excited state n to the ground state. The wavelength so emitted indicates a photosensitive material having a work function of 2.75 eV. If the stopping potential of the photoelectron is 10V, then the value of n is

According to Einstein’s photoelectric equation

eV₀ = hf- W₀

Or, hf – eV₀ +W₀

= 10+ 2.75

= 12.75 eV

When an electron In the hydrogen atom makes a transition from excited state n to the ground state (n = 1), then the frequency (f) of the emitted photon is given by,

hf = \(E_n-E_1\)

=\(-\frac{13.6}{n^2}-\left(-\frac{13.6}{1^2}\right)\)

Or, n= 4

Question 37. Out of the following which one is not a possible energy for a photon to be emitted by a hydrogen atom according to Bohr’s atomic model?

1. 0.65 eV
2. 1.9 eV
3. 11.1 eV
4. 13.6 eV

Answer: 3. 11.1 eV

Hint: The energy of n th orbit of a hydrogen atom is given by,

⇒ \(E_n=-\frac{13.6}{n^2}\)

∴ E_∞ – E₁ = 0- (13.6 ) = 13.6 eV

E₃ – E₂ = = -1.5 -(-3.4) = 1.9 eV

E₄ – E₃ = 0.85 – (1.5 ) eV =  0.65 eV

Class 12 physics dual nature questions 

Question 38. The wavelength λ and the frequency f of a particular X-ray spectral line varies with the atomic number Z of the target element. In this event, Z is nearly proportional to

1. λ
2. √λ
3. f
4. √f

Answer: 4. √f

Question 39. The ratio of the wavelengths of Ka and KQ spectral lines of hydrogen is

1. 8: 27
2. 16: 27
3. 27: 32
4. 9: 16

Answer: 3. 27: 32

Question 40. Let, λ₂₁, and λ₃₁ and λ₃₂ be the wavelengths of the spectral lines for the transition of electrons in the energy levels 2 → 1, 3 → 1, and 3 → 2 of an atom. Then

1.  λ₂₁ > λ₃₁
2.  λ₂₁ > λ₃₂
3.  λ₃₁ > λ₃₂
4.  λ₃₁ > λ₃₂

Answer: 1 and 3

Question 41. Energy in the second energy state of a hydrogen atom is E₂ Then

1. Energy in the third energy state of He⁺ ion (Z =.2) is
2. Energy in the ground state of He⁺ ion (Z = 2) is 16E₂
3. Energy in the third energy state of Li²⁺ ion (Z = 3) is
4. Energy in the second energy state of Li2+ ion (Z = 3) is 9E₂

Answer: 2, 3, and 4

Question 42. The radius of the orbit of an electron in the ground state of a hydrogen atom is a₀. Then

1. The radius of the orbit of an electron in the second energy state of the He⁺ ion (Z = 2) is 2a₀
2. The radius of orbit of an electron in the third energy state of He+ ion (Z = 2) is 3a₀
3. The radius of orbit of an electron in the ground state of Li²⁺ ion (Z = 3) is a₀/3
4. The radius of orbit of an electron in the third energy state of Li²⁺ ion (Z = 3) is 3a₀

Answer: 1, 3 and 4

Question 43. In the hydrogen atom spectrum

1. Lines of the Balmer series are in the visible region
2. Lines of the Lyman series are in the ultraviolet region
3. Lines of the Paschen series are in the ultraviolet region
4. Lines of Brackett series arc in the infrared region

Answer: 1, 2 and 3

Class 12 physics dual nature questions 

Question 44. In the case of X-ray spectrum

1. Cut-off wavelength depends on the kinetic energy of the electrons incident on the target
2. Cut-off wavelength depends on. the material of the target
3. Wavelength of Ka -line depends on the material of the target
4. The Wavelength of Kg -line is larger than the wavelength of

Answer: 1 and 3

Question 45. H, He⁺, and Li²⁺ are examples of .atoms or ions with one electron each. The energy of such atoms when in the n-th energy state (according to Bohr’s theory, n = 1, 2, 3,……. = principal quantum number) is E_n = \(-\frac{13.6 \mathrm{Z}^2}{n^2}\)  (1eV = 1.6 × 10^-19 J). For the ground state, n = 1 . To raise the atom from the ground state to n = f, the suitable incident light should have a wavelength given by λ  = \(\frac{h c}{E_f-E_1}\) But the atom cannot stay permanently in the E_f – E₁, energy state, ultimately, it comes to the ground state by radiating extra energy, E_f – E₁, as electromagnetic radiation. The electron of the atom comes from n = f to n = 1 in one or more steps using the permitted energy levels. As a result, there is a possibility of emission of radia¬tion with more than one wavelength from the atom. Planck’s constant = 6.63 × 10^-34 J.s. s and velocity of light c = 3× 10⁸ m.s^-1

1. What Is the wavelength of the light incident on thr atom to raise it to the fourth quantum level hum ground stale?

1. 952 Å
2. 975 Å
3. 1027 Å
4. 1219 Å

Answer:  2. 975 Å

2. Radiations of how many wavelengths are possible in case of die excited atom In example 1 to come to ground state?

1. 2
2. 3
3. 6
4. 9

Answer: 3. 6

3.  What is the value of the maximum wavelength in example

1. 952  Å
2. 975 Å
3. 577 Å
4. 10630 Å

Answer:   4. 10630 Å

4. What is the value of the minimum wavelength in example 

1. 952 Å
2. 975 Å
3. 6577 Å
4. 18830 Å

Answer: 2. 975 Å

Class 12 physics dual nature questions 

5. Energy of which quantum state of He⁺ ion will be equal to the ground-level energy of hydrogen?

1. n = 1
2. n = 2
3. n = 3
4.  n = 4

Answer:  2. n= 2

6. The wavelength of radiation emitted for the transition of the electron of He4 ion from n = 1 to n & 2 Is

1. 952 Å
2. 975 Å
3. 1027 Å
4. 1219 Å

Answer: 4. 1219 Å

7. For what wavelength of Incident radiation He⁺ ion will be raised to the fourth quantum state from the ground state?

1. 243.7 Å
2. 487.5 Å
3. 731.2 Å
4. 975 Å

Answer:  1. 243.7 Å

8. Which among the following differences In the energy levels for a Li²⁺ Ion is minimum?

1. E₂– E₁
2. E₃ – E₂
3. E₃ – E₁
4. E₄ – E₃

Answer:  4. E₄ – E₃

Real-Life Applications of Atomic Concepts

Question 46. The total energy of an electron for any particular energy level in a hydrogen atom Is -1,51 eV. The value of the principle quantum number of the energy level is

1. n = 2
2. n = 1
3. n = 3
4. n = 4

Answer: 3. n = 3

The energy of electron In n -th energy level, wavelength

⇒ \(E_n=-\frac{13.6}{n^2}\)

-1. 51 = \(-\frac{13.6}{n^2}\)

Or, n = \(\sqrt{\frac{13.6}{1.51}}\)

= 3

Question 47. The ratio of the minimum wavelength of Lyman and Balmer scries In the hydrogen spectrum will be

1. 10
2. 5
3. 0.25
4. 1.25

Answer: 3. 0.25

Question 48. If V is the accelerating voltage, then the maximum frequency of X-ray emitted from an X-ray tube is

1. \(\frac{e h}{V}\)
2. \(\frac{e V}{h}\)
3. \(\frac{h}{e V}\)
4. None of these

Answer: 2. \(\frac{e V}{h}\)

The energy of an electron = eV = Iwmax = maximum energy of an X-ray photon

⇒ \(\nu_{\max }=\frac{e V}{h}\)

Photoelectric effect multiple choice questions 

Question 49. The ionization energy of hydrogen is 13.6 eV. The energy of tire photon released when an electron jumps from the first excited state (n = 2) to the ground state of a hydrogen atom is

1. 3.4 eV
2. 4.53 eV
3. 10.2 eV
4. 13.6 eV

Answer: 3. 10.2 eV

The energy of the released photon

= \(13.6\left[\frac{1}{n_i^2}-\frac{1}{n_f^2}\right]\)

= \(13.6\left[\frac{1}{1^2}-\frac{1}{2^2}\right]\)

= 10.2 eV

Question 50. A photon of wavelength 300 nm interacts with a stationary hydrogen atom in the ground state. During the interaction, the whole energy of the tire photon is transferred to the electron of the atom. State which possibility is correct considering Planck’s constant = 4 × 40^-15eV velocity of light = 3 × 10⁸  m/s, the ionization energy of hydrogen =13.6 eV.

1. The election will be knocked out of the atom
2. Electrons will go to any excited seas of dead atom
3. The electron will go only to the first excited state of the atom
4. The electron will keep orbiting in the ground state of an atom

Answer: 4. Electron will keep orbiting in the ground state of

Energy ofthe photon

E = \(hf\frac{h c}{\lambda}=\frac{\left(4 \times 10^{-15}\right) \times\left(3 \times 10^6\right)}{300 \times 10^{-9}}\)

= 4 eV

Groundstate energy of hydrogen atom = – 13.6eV

Energy ofthe second orbit = \(\frac{13.6}{2^2}\)

= -3.4 eV

∴ The energy required to move the electron from the ground state second orbit = -3.4- (-13.6) = 10.2 eV.

Hence, the electron will keep orbiting the groundstate of the atom

Question 51. The number of Broglie wavelengths contained in the second Bohr orbit of a Hydrogen atom is
Answer:

1. 1
2. 2
3. 3
4. 4

Answer: 2. 2

Question 52. The wavelength of the second Balmer line in the Hydrogen spectrum is 600 nm. The wavelength for its third line in the Lyntann series is
Answer:

1. 800
2. 600
3. 400
4. None of these

Answer:  4. None of these

We know, for the second line in the Balmer series.

⇒ \(\frac{1}{\lambda_1}=R\left[\frac{1}{4}-\frac{1}{16}\right]=\frac{3 R}{16}\)

For tliini line in Lyniann series,

⇒  \(\frac{1}{\lambda_2}=R\left[\frac{1}{1}-\frac{1}{16}\right]=\frac{15}{16} R\)

⇒  \(\frac{\lambda_2}{\lambda_1}=\frac{3 R}{16} \times \frac{16}{15 R}=\frac{1}{5}\)

Or, \(\lambda_2=\frac{1}{5} \times \lambda_1=\frac{1}{5} \times 600 \mathrm{~nm}\)

= 120 nm

Photoelectric effect multiple choice questions 

Question 53. Let vn and En be the respective speed and energy of an electron in the -th orbit of radius r_n, in a hydrogen atom, as predicted by Bohr’s model. Then

1. The plot of \(E_n r_n / E_1 r_1\) as a function of n is a straight line of slope 0
2. The plot of \(r_n v_n / r_1 v_1\) as a function of n is a straight line of slope 1
3. Plot of In \(\left(\frac{r_n}{r_1}\right)\) as a function of In(n) is a straight line of slope 2
4. Plot of \(\left(\frac{r_n E_1}{E_n r_1}\right) \) as a function of In(n) is a straight line of slope 4

Answer:  1,2, 3 and 4

According to Bohr’s theory, vn \(v_n \propto \frac{1}{n}\) ……………. (1)

\(E_n \propto \frac{1}{n^2}\) ……………. (2)

And \(r_n \propto n^2\) ……………. (3)

From equations (2) and (3) we get

⇒ \(E_n r_n \propto n^2 \times \frac{1}{n^2} \text { or, } E_n r_n \propto n^0\)

Hence En rn = constant

⇒ \(\frac{E_n r_n}{E_1 r_1}\)

From equations (1) and (3) we get, rnt/n« n

⇒ \(\frac{r_n v_n}{r_1 v_1}=n \text { or, } \frac{\left(r_n v_n\right) /\left(r_1 v_1\right)}{n}\)

= 1

From equation (3) \(\frac{r_n}{E_n} \propto n^4\)

⇒ \(\frac{r_n E_1}{E_n r_1}=n^4 \text { or, } \frac{\ln \left(r_n E_1 / E_n r_1\right)}{\ln (n)}\) = 4

Question 54. How the linear velocity v of an electron in the Bohr orbit Is related to its quantum number n?

1. \(v \propto \frac{1}{n}\)
2. \(v \propto \frac{1}{n^2}\)
3. \(v \propto \frac{1}{\sqrt{n}}\)
4. \(\nu \propto n\)

Answer: 1. \(v \propto \frac{1}{n}\)

We know, for n -th orbit

⇒ \(m v_n r_n=\frac{n h}{2 \pi}\)

Or, \(v_n=\frac{n h}{2 \pi m r_n}\)

⇒ \(v_n \propto \frac{1}{n}\left[r_n=\frac{\epsilon_0 n^2 h^2}{\pi m Z e^2}\right]\)

Question 55. The radiation corresponding to the 3 → 2 transition of hydrogen atom falls on a metal surface to produce photoelectrons. These electrons are made to enter a magnetic field of 3 × 10^-4  T.If the radius of the largest circular path followed by these electrons is 10.0 mm, the work function ofthe metal is close to

1. 1.6 eV
2. 1.8 eV
3. 1.1 eV
4. 0.8 eV

Answer: 3. 1.1 eV

The kinetic energy of an electron

= \(E_K=\frac{q^2 B^2 R^2}{2 m}\)

= \(=\frac{\left(1.6 \times 10^{-19}\right)^2 \times\left(3 \times 10^{-4}\right)^2 \times\left(10 \times 10^{-3}\right)^2}{2 \times 9.1 \times 10^{-31} \times 1.6 \times 10^{-19}}\)

= 0.79 eV

For the transition of an electron from energy state 3 to 2

E = \(13.6\left(\frac{1}{2^2}-\frac{1}{3^2}\right)=\frac{13.6 \times 5}{36}\)

= 1.88 eV

According to Einstein’s photoelectric equation

E = E_K+ W₀

Or, W₀ = E- E_k

= 1.88 – 0.79 ≈ 1.1 ev

Question 56. Hydrogen (¹H₁), deuterium (²H₁)), singly ionized helium (⁴He₂)^+, and doubly ionized lithium (⁶Li₃)⁺⁺ all have electrons around the nucleus. Consider an electron transition from n = 2 to n = 1. If the wavelengths of emitted radiation are, λ₁, λ₂, λ₃ and λ₄ respectively then approximately which one of the following is correct?

1. λ₁ = 2λ₂= 3λ₃= 4λ₄
2. 4λ₁= 2λ₂=2λ₂ = λ₄
3. λ₁ = 2λ₂ = 2λ₃, = λ₄
4. λ₁= λ₂= 4λ₃= 9λ₄

Answer: 4. λ₁= λ₂= 4λ₃= 9λ₄

For hydrogen, Z = 1 , for deuterium, Z = 1

For helium ion, Z = 2 ; for lithium-ion, Z = 3

Photoelectric effect multiple choice questions 

Question 57. As an electron makes a transition from an excited state to the ground state of a hydrogen-like atom ion

1. Its kinetic energy increases but potential energy and total energy decrease
2. Kinetic energy, potential energy, and total energy decrease
3. Kinetic energy decreases, potential energy increases but total energy remains the same
4. Kinetic energy and total energy decrease but potential

Answer:  1. Its kinetic energy increases but potential energy and total energy decrease

The kinetic energy, potential energy, and total energy of a revolving electron in an atom or ion are E_k, E_p, and E respectively.

E_K + E_p = E-,E_K = -E-,E_p= 2E

∴ E_K is always positive,

∴  E_p and E are always negative

E gets decreased as an electron makes a transition from an excited state to the ground state of a hydrogen-like atom/ ion.

Hence, E_p also decreases, but E_K increases

Question 58. An electron beam is accelerated by a potential difference V to hit a metallic target to produce X-rays. It produces continuous as well as characteristic X-rays. If λ_min  is the smallest possible wavelength of X-ray in the spectrum, the variation of log λ_min with IogV is correctly represented in

Answer:  1

⇒ \(\lambda_{\min }=\frac{h c}{e V}\)

Or, \(\log \lambda_{\min }=-\log V+\log \left(\frac{h c}{e}\right)\)

The negative sign implies a negative slope

Question 59. Some energy levels of a molecule are. The ratio ofthe wavelengths r = λ₁/ λ₂ is given by

1. \(r=\frac{4}{3}\)
2. \(r=\frac{2}{3}\)
3. \(r=\frac{3}{4}\)
4. \(r=\frac{1}{3}\)

Answer: 4. \(r=\frac{1}{3}\)

According to Bohr’s postulate

ΔE = hν = \(h \frac{c}{\lambda}\)

Or, \(\lambda=\frac{h c}{\Delta E}\)

In this case \(\frac{\lambda_1}{\lambda_2}=\frac{\Delta E_2}{\Delta E_1}=\frac{\left(-\frac{4}{3} E\right)-(-E)}{(-2 E)-(-E)}\)

⇒ \(\frac{1}{3}\)

Examples of Spectral Series Questions

Question 60. If the series  limit frequency of the Lyman series is vL, then the series limit frequency of the Pfund series is

1. \(\frac{\nu_L}{16}\)
2. \(\frac{\nu_L}{25}\)
3. 25ν_L
4. 16ν_L

Answer: 2. \(\frac{\nu_L}{25}\)

For the series limit frequency of the Lyman series

\(h \nu_L=E\left[\frac{1}{1^2}-\frac{1}{\infty}\right]\) = E

Again, for the series limit frequency of the Pfund series,

⇒ \(h \nu_p=E\left[\frac{1}{5^2}-\frac{1}{\infty}\right]=\frac{E}{25}\)

⇒ \(\nu_P=\frac{\nu_L}{25}\)

WBCHSE physics dual nature MCQs 

Question 61. An electron from various excited states of hydrogen atom emit radiation to come to the ground state. Let An, A& be the de Broglie wavelength of the electron in the n -th state and the groundstate respectively. Let An be the wavelength of the emitted photon in the transition from the n -th state to the ground state. For large n,[A, B are constants)

1. \(\Lambda_n^2 \approx A+B \lambda_n^2\)
2. \(\Lambda_n^2 \approx \lambda\)
3. \(\Lambda_n \approx A+\frac{B}{\lambda_n^2}\)
4. \(\Lambda_n \approx A+B \lambda_n\)

Answer: 3. \(\Lambda_n \approx A+\frac{B}{\lambda_n^2}\)

We Know,

Question 62. A hydrogen atom in the ground state is excited by a monochromatic radiation of = 975 Å. The number of spectral lines in the resulting spectrum emitted will be

1. 3
2. 2
3. 6
4. 10

Answer: 3. 6

From Rydberg formula

⇒ \(\frac{1}{\lambda}=R\left(\frac{1}{1^2}-\frac{1}{n^2}\right)\)

⇒  \(\left(1-\frac{1}{n^2}\right)=\frac{1}{\lambda R}=\frac{1}{975 \times 10^{-8} \times 109706}\)

= 0.334

⇒  \(\frac{1}{n^2}=1-0.934\)

= 0.065

Or, n = 15.36

or, n = 3.91 ≈ 4

∴ Number of line spectra = ⁿC₂ = ⁴C₂

= 6

Question 63. Consider the 3rd orbit of He⁺ (Helium), using the non-relativistic approach, the speed of an electron in this orbit will be [given k = 9 × 10⁹ constants, Z = 2 and h (Planck’s constant) = 6.6 × 10^-34  J.s ]

1. 2.92 × 10⁶ m/s
2. 0.73 × 10⁶ m/s
3. 1.46 × 10⁶ m/s
4. 3.0 × 10⁸ m/s

Answer: 2. 0.73 × 10⁶ m/s

According to Bohr’s quantum condition

⇒ \(m v_n r_n=n \frac{h}{2 \pi} \text { or, } v_n=\frac{n h}{2 \pi m r_n}\)

Again \(r_n \propto \frac{n^2}{m Z} \text { or, } m r_n \propto \frac{n^2}{Z}\)

Hence \(v_n \propto n \frac{Z}{n^2} \quad \text { i.e., } v_n \propto \frac{Z}{n}\)

For ground state of hydrogen, Z = 1 and n = 1

⇒  \(v=\frac{c}{137}\)

For rhird orbit of he Z= 2

⇒  \(\frac{c}{137} \cdot \frac{2}{3}=\frac{3 \times 10^8}{137} \times \frac{2}{3}\)

Conceptual Questions on Electron Configuration

Question 64. Given the value of the Rydberg constant is 10⁶ m^-1,  wave number of the last line of the Balmer series in the hydrogen spectrum will be

1. 0.53× 10⁷ m^-1
2. 0.25 × 10⁷ m^-1
3. 2.5 × 10⁷ m^-1
4. 0.025 × 10⁴ m^-1

Answer: 2. 0.25 × 10⁷ m^-1

⇒  \(\frac{1}{\lambda}=R\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)\)

The wave number ofthe last line ofthe Balmer series,

⇒  \(\bar{\nu}=\frac{1}{\lambda}=10^7\left(\frac{1}{2^2}-\frac{1}{\infty^2}\right)\)

= 0.25 × 10⁷ m^-1

Question 65. If the longest wavelength in the ultraviolet region of the hydrogen spectrum is the shortest wavelength in its infrared region is

1. \(\frac{46}{7} \lambda_0\)
2. \(\frac{20}{3} \lambda_0\)
3. \(\frac{36}{5} \lambda_0\)
4. \(\frac{27}{4} \lambda_0\)

Answer: 4. \(\frac{27}{4} \lambda_0\)

The longest wavelength in the ultraviolet region of

⇒ \(\frac{1}{\lambda_0}=R\left(\frac{1}{1^2}-\frac{1}{2^2}\right)\)

= \(\frac{3}{4}\) R

Or, \(\lambda_0=\frac{4}{3 R}\)

Again, if the shortest wavelength in the infrared region of the hydrogen spectrum = the shortest wavelength in the Paschen series = A

⇒ \(\frac{1}{\lambda}=R\left(\frac{1}{3^2}-\frac{1}{\infty^2}\right)=\frac{R}{9}\)

Or, \(\lambda=\frac{9}{R}=\frac{27}{4} \lambda_0\)

WBCHSE physics dual nature MCQs 

Question 66. The ratio of kinetic energy to the total energy of an electron in a Bohr orbit ofthe hydrogen atom is

1. 2: -1
2. 1: – 1
3. 1:1
4. 1:-2

Answer: 2. 1: – 1

The total energy and the kinetic energy of an electron in n nth Bohr radius is E_n and E_k respectively

E_k = – E_n

Or, \(\frac{E_K}{E_n}=1:(-1)\)

WBCHSE Class 12 Physics MCQs

Dual Nature Of Matter And Radiation Multiple Questions And Answers

Question 1. A particle is dropped from a height of H. The de Broglie wavelength of the particle as a function of height is proportional to

1. H
2. H^½
3. H
4. H^-½

Answer: 4. H^-½

Question 2. Consider a beam of electrons (each electron with energy E₀) incident on a metal surface kept in an evacuated chamber. Then

1. No electrons will be emitted as only photons can emit electrons
2. Electrons can be emitted but all with an energy E_p
3. Electrons can be emitted with any energy, with a maximum of E_p– Φ (Φ is the work function)
4. Electrons can be emitted with any energy, with a maximum of E₀

Answer: 4. Electrons can be emitted with any energy, with a maximum of E₀

Question 3. An electron (mass m ) with an \(\vec{V}=v_0 \hat{i}\) is in an electric (\(\vec{E}=E_0 \hat{j}\)  constant E₀> 0 ). Its de Broglie wavelength at time t is given by

1. \(\frac{\lambda_0}{\left(1+\frac{e E_0}{m} \cdot \frac{t}{v_0}\right)}\)
2. \(\lambda_0\left(1+\frac{e E_0 t}{m v_0}\right)\)
3. λ₀
4. λ₀t

Answer: 1. \(\frac{\lambda_0}{\left(1+\frac{e E_0}{m} \cdot \frac{t}{v_0}\right)}\)

Question 4. An electron (mass m ) with an initial velocity \(\vec{V}=v_0 \hat{i}\)  is in an electric field \(\vec{E}=E_0 \hat{j}\). \(\lambda_0=\frac{h}{m v_0}\)  If,  its de Broglie wavelength at time t is given by

1. λ₀
2. \(\lambda_0 \sqrt{1+\frac{e^2 E_0^2 t^2}{m_0^2 v_0^2}}\)
3. \(\frac{\lambda_0}{\sqrt{1+\frac{e^2 E_0^2 t^2}{m_0^2 \nu_0^2}}}\)
4. \(\frac{\lambda_0}{\left(1+\frac{e^2 E_0^2 t^2}{m_0^2 v_0^2}\right)}\)

Answer: 3. \(\frac{\lambda_0}{\sqrt{1+\frac{e^2 E_0^2 t^2}{m_0^2 \nu_0^2}}}\)

Read and Learn More Class 12 Physics Multiple Choice Questions

Question 5. An electron is moving with an initial velocity \(\vec{V}=v_0 \hat{i}\)  and is in a magnetic field \(\vec{B}=B_0 \hat{i}\),  its de Broglie wavelength

1. Remains constant
2. Increase with time
3. Decreases with time
4. Increases and decreases periodically

Answer: 1. Remains constant

Dual Nature of Matter Short Questions WBCHSE

Question 6. Relativistic corrections become necessary when the expression for kinetic energy ½mv² becomes comparable to me2. At what de Broglie wavelength will relativistic corrections become important for an electron?

1. 10 nm
2. 10 nm
3. 10 nm
4. 10-4 nm

Answer: 3 and 4

Question 7. Photons absorbed in matter are converted to heat. A source emitting n photons of frequency f is used to convert 1 kg of ice at 0°C to water at 0°C. Then the time T taken for the conversion

1. Decreases with increasing n, with f fixed
2. Decreases with n fixed, f Increasing
3. Remains constant with n and f changing such that nf= constant
4. Increases when product nf increases

Answer: 1, 2, And 3

WBCHSE class 12 physics MCQs 

Question 8. A particle moves in a closed orbit around the origin, due to a force that is directed towards the origin. The de Broglie wavelength of the particle varies cyclically between two values, λ₁,λ₂ with  λ₁>λ₂ Which of the following statements is true?

1. The particle could be moving in a circular orbit with the origin as the center.
2. The particle could be moving in an elliptic orbit with the origin as the focus.
3. When the de Broglie wavelength is λ₁, the particle is the origin than when its value is λ₂.
4. When the de Broglie wavelength is λ₂, the particle is nearer the origin than when its value is λ₂

Question 9. A monochromatic source of light is kept at a distance of 0.2 m from a photoelectric cell. Stopping potential V0 and sat¬ uration current I0 are 0.6V and 18.0 mA, respectively. Now the source is kept at a distance of 0.6m from the cell. Then

1. V₀ = 0.2V, I₀= 18.0mA
2. V₀ = 0.2 V, I₀= 2.0mA
3. V₀ = o.6V,I₀ = 18.0mA
4. V₀ = 0.6V, I₀ = 2.0mA

Answer: 4. V₀ = 0.6V, I₀ = 2.0mA

Question 10. For a monochromatic light incident on a metal surface, the stopping potential is V. Then the kinetic energy of the fast¬ est photoelectrons emitted from that surface is

1. eV
2. 2eV
3. \(\frac{2 e V}{m}\)
4. \(\sqrt{\frac{2 e V}{m}}\)

Answer: eV

WBCHSE Physics Questions on Dual Nature of Radiation

Question 11. If in a photo-electric experiment, the wavelength of inci¬ dent radiation is reduced from 6000 A to 4000 A then

1. Stopping potential will decrease
2. Stopping potential will increase
3. The kinetic energy of emitted electrons will decrease
4. The value of the work function will decrease

Answer: 2. Stopping potential will increase

Question 12. If the stopping potential for photoelectric emission is 0.75 V,’ the kinetic energy of the fastest photoelectrons is

1. 0.75 V
2. 7.5eV
3. 0. 7.5eV
4. 0.75 × 10^-19 J

Answer: 3. 0. 7.5eV

Question 13. For a monochromatic light incident on a metal surface, the maximum velocity of the emitted photoelectrons is v. Then the stopping potential would be

1. \(\frac{2 m v^2}{e}\)
2. \(\frac{m v^2}{e}\)
3. \(\frac{m v^2}{2e}\)
4. \(\frac{m v^2}{\sqrt{2} e}\)

Answer: 3. \(\frac{m v^2}{2e}\)

Question 14. For two monochromatic radiations incident on the same metal surface, the stopping potentials are 1.0 V and 2.0V. The ratio between the maximum velocities, of the emitted photoelectrons is

1. 2:1
2. √2:1
3. 1:√2
4. 1:2

Answer: 3. 1: √2

WBCHSE class 12 physics MCQs 

Question 15. The energy of photon incident on a metal plate is twice its work function. How many times should be the wavelength of incident light so that the kinetic energy of the fastest elec¬ tron will be doubled?

1. \(\frac{3}{2}\) times
2. \(\frac{2}{3}\) times
3. \(\frac{1}{2}\) times
4. 2 times

Answer: 2. \(\frac{2}{3}\) times

Brief Q&A on Dual Nature of Matter for Students

Question 16. The work function of zinc is twice that of sodium. If the photo¬ electric threshold wavelength for sodium is 7000A, what will be its value for zinc?

1. 3500A°
2. 14000A°
3. 10500A°
4. 4667A°

Answer: 1. 3500A°

Question 17. The threshold wavelength of a metal for electron emission is 5200A. Which one of the following sources of light will be able to emit electrons from the metal?

1. 50 W infrared
2. 1W infrared
3. 50 W red light
4. 1W ultraviolet

Answer: 4. 1W ultraviolet

Question 18. When photons of energy 6eV are incident on a metal surface, the kinetic energy of the fastest electrons becomes 4eV. The value of stopping potential is (in V)

1. 2
2. 4
3. 6
4. 10

Answer: 2 . 4

Dual nature of matter and radiation class 12 MCQs 

Question 19. In the case of the photoelectric effect, the incident photon

1. Vanishes completely
2. Is scattered with lower frequency
3. Is scattered with higher frequency
4. Is scattered with the same frequency

Answer: 1. Vanishes completely

Question 20. Which of the following quantities has the same dimension as that of Planck’s constant?

1. Linear momentum
2. Angular momentum
3. Energy
4. Power

Answer: 2. Angular momentum

Question 21. In the case of the photoelectric effect, the graph of the kinetic energy of the photoelectron concerning the frequency of incident radiation will be a straight line. The slope of this straight line depends on

1. The nature of the metal surface
2. The intensity of the incident radiation
3. The nature of the metal surface as well as the intensity of incident radiation
4. None of the nature of the metal surface or the intensity

Answer: 4. None of the nature of the metal surface or the intensity

Essential Questions About Dual Nature of Matter WBCHSE

Question 22. A body absorbs 5 x 1029 photons of frequency 102° Hz. Which of the following information is correct? [Assume, all the energy of photons is transformed into mass.]

1. The mass of the body remains unchanged
2. The mass of the body increases by 0.00037 kg
3. The mass of the body increases by 0.37 kg
4. The mass of the body increases by 3.7 kg

Answer: 3. Mass of the body increases by 0.37 kg

Question 23. The momentum of a photon (frequency = f, rest mass = 0 )

1. \(\frac{h f}{c}\)
2. \(\frac{h \lambda}{c}\)
3. \(\frac{h c}{\lambda}\)
4.  Zero

Answer: 1. \(\frac{h f}{c}\)

Dual nature of matter and radiation class 12 MCQs 

Question 24. A monochromatic radiation of wavelength A and intensity f is incident on a plate of area A. Find the number of photons striking the plate per second.

1. \(\frac{I \lambda}{A h c}\)
2. \(\frac{h c}{I \lambda A}\)
3. \(\frac{I \lambda A}{h c}\)
4. \(\frac{h \lambda}{c \lambda A}\)

Answer: 3. \(\frac{I \lambda A}{h c}\)

Question 25. The threshold frequency for a photosensitive metal is 3.3 × 10¹⁴ Hz. If a light of frequency 8.2 × 10¹⁴Hz is incident on this metal, the cut-off voltage for the photoelectron emission is nearly

1. 1V
2. 2V
3. 3V
4. 5V

Answer: 2. 2V

Question 26. The anode voltage of n photocell I kept fixed, The 24. The wave wavelength λ of the light falling on the cathode Is gradually changed. The plate current I of the photocell varies as follows

Answer: 2

Dual nature of matter and radiation class 12 MCQs 

Question 27. A body of mass 60g is moving with a velocity of 10 m. s^-1. The de Broglie wavelength of the body will be approximately (h = 6.63 × 10^-34J.s )

1. 10^-35m
2. 10^-25 m
3. 10^-33m
4. 10^-23m

Answer: 3. 10^-33m

Question 28. If the de Broglie wavelength of a gas molecule at 0°C is λ, what will be its wavelength at 819°C?

1. λ
2. λ/2
3. λ/3
4. λ/4

Answer:  2. λ/2

Question 29. Two moving electrons have a ratio of 1: 2 between their respective kinetic energies. The ratio between their de Broglie wavelength is

1. 2:1
2. √2:1
3. 1: √2
4. 1:2

Answer:  2. √2:1

Class 12 physics dual nature questions 

Question 30. An electron is accelerated by a potential difference V and as a result, its de Broglie wavelength becomes A. If the applied potential difference was 2 V, the de Broglie wave¬ length would have been

1. 2λ
2. √2λ
3. λ/√2
4. λ/2

Answer: 4. λ/2

Question 31. The wavelength associated with an electron of mass m having kinetic energy E is given by

1. \(\frac{2 h}{m E}\)
2. 2mhE
3. \(\frac{2 \sqrt{2 m E}}{h}\)
4. \(\frac{h}{\sqrt{2 m E}}\)

Answer: 4. \(\frac{h}{\sqrt{2 m E}}\)

Question 32. The wavelength of a light is. 0.01 A° .If h is Planck constant, then the momentum of the corresponding photon will be

1. 10^-2 h
2. h
3. 10²h
4. 10¹² h

Answer: 4. 10¹² h

Question 33. Electrons used in an electron microscope are accelerated by a voltage of 25 kV. If the voltage is increased to 100kV then the de Broglie wavelength associated with the electrons would

1. Increase by 2 times
2. Decrease by 2 times
3. Decrease by 4 times
4. Increase by 4 times

Answer: 2. Decrease by 2 times

Class 12 physics dual nature questions 

Question 34. The de Broglie wavelength associated with proton changes by 0.25% If its momentum is changed by pQ. The initial momentum was

1. 100 p₀
2. p₀ / 400
3. 401 p₀
4. p₀ /100

Answer: 3. 401 p₀

Question 35. The proton when accelerated through a potential difference of V volt has a wavelength A associated with it. If an a -particle is to have the same wavelength A, it must be accelerated through a potential difference of

1. \(\frac{V}{8}\)
2. \(\frac{V}{4}\)
3. 4V
4. 8V

Answer: 1. \(\)

Class 11 Physics	Class 12 Maths	Class 11 Chemistry
NEET Foundation	Class 12 Physics	NEET Physics

Question 36. The threshold frequency of a photoelectric effect depends on

1. Nature of the metal surface
2. The intensity of the incident radiation
3. The energy of the incident photon
4. The work function of the metal

Answer: 1 And 4

Question 37. Maximum kinetic energy of photoelectron depends on

1. Nature of the metal surface
2. Intensity of incident radiation
3. The energy of an incident photon
4. The work function of the metal

Answer: 1,3 And 4

Photoelectric effect multiple choice questions 

Question 38. The work function of a metal surface is 2.0 eV. Light of wavelength 5000 A is incident on it:

1. The energy of each incident photon is 2.48 eV
2. The threshold wavelength for the photoelectric effect is 6200 A°
3. The maximum kinetic energy of the emitted photoelectron is 0.48 eV
4. The stopping potential is 0.48 eV

Answer: 1,2,3 And 4

Question 39. n number of photons of frequency f are emitted per second from a light source of power P (h =Planck’s constant; c = speed of light). Then

1. n = \(\frac{P}{h f}\)
2. Energy of each photon = \(\frac{P}{c n}\)
3. Momentum of each photon = \(\frac{P}{p n}\)
4. The value of n increases if the wavelength of the light increases

Answer: 1,3 And 4

Question 40. The threshold frequency and the threshold wavelength of photoelectric emission from a metal surface are fQ and. The; frequency and the wavelength of incident light are f and λ₀. Then

1. There will be no photoelectric effect If f>f₀
2. There will be no photoelectric effect If  λ> λ₀
3. Stopping potential ∝ ( f – f₀)
4. Maximum kinetic energy of photoelectron ∝ (f>f₀)

Answer: 2 And 3

Question 41. Work functions of two metals .1 and It art’ 3.1 eV and 1.9 eV respectively light of wavelength 3000 A° Is incident on both the surfaces.

1. No photo mission will take place in case of metal A
2. Photoelectrons will be emitted from both metals
3. The maximum kinetic energy of the photoelectron will be higher in metal It
4. Threshold wavelength of photoelectric effect In the case of metal .-1 will lie <1000 A° approximately

Answer: 2,3 And 4

Question 42. The de Broglie wavelength of a moving particle of mass m is A . For a few particles of different masses

1. \(\lambda \propto \frac{1}{m}\) , If their momenta are same
2. \(\lambda \propto \frac{1}{m}\) , If their velocities are same
3. \(\lambda \propto \frac{1}{m}\) , If kinetic energies are same
4. \(\lambda \propto \frac{1}{\sqrt{m}}\), If their kinetic energies are same

Answer: 2 And 4

Photoelectric effect multiple choice questions 

Question 43. An electron (mass in ) and a proton (mass M) are acceler¬ rated with the same potential difference, then

1. Ratio of their velocities = \(\sqrt{\frac{M}{m}}\)
2. Ratio of their momenta = \(\sqrt{\frac{M}{m}}\)
3. The ratio of their kinetic energies = 1
4. Ratio of their de Broglie wavelengths = \(\sqrt{\frac{M}{m}}\)

Answer: 1, 3 And 4

Question 44. The wavelength of Ka X-ray for lead isotopes Pb208, Ph206, Pb204 are Av A2 and A3 respectively. Then

1. λ₁ = λ₂ =  λ₃
2. λ₁ > λ₂ > λ₃
3. λ₁ < λ₂ <  λ₃
4. \(\sqrt{\lambda_1 \lambda_2}\)

Answer: 1 And 4

Question 45. In which of the following situations the heavier of the two particles have a smaller de Broglie wavelength? The two particles

1. Move with the same speed
2. Move with the same kinetic energy
3. Move with the same linear momentum
4. Have fallen through the same height

Answer: 1,2 And 4

Question 46. The graph represents the variation of maximum kinetic energy with the frequency of an emitted photoelectron. this graph helps us to determine.

1. Work function
2. Planks constant
3. Threshold frequency
4. Charge on an electron

Answer: 1,2 And 3

Question 47. Einstein established the idea of photons or the basis of Planck’s quantum theory. According to his idea? the light of frequency f or wavelength A is a stream of photons The rest mass of each photon is zero velocity is equal to the mass of each photon, and velocity is equal to the velocity of light (c)
= 3 ×10^-8, J.s energy E = hf , where h = planks constant = 6.625 ×10^-34 J.s . Each photon has a momentum p= \(\), although its rest mass is zero. The number of photons increases when the intensity of incident light increases and vice-versa

On the other hand, according to de Broglie, any stream of moving particles may be represented by progressive waves. The wavelength of the wave (de Broglie wavelength) is λ = h/p, where p is the momentum of the particle. When a particle having charge e is accelerated with a potential difference of V, the kinetic energy gained by the particle is K = eV. Thus as the applied potential difference is increased, the kinetic energy of the particle and hence the momentum increase resulting in a decrease in the de Broglie wavelength. Given, a charge of electron e = 1.6 ×10^-19C and mass = 9.1×10^-31kg.

1. The energy (in eV) of each photon associated with the light of wavelength 5893 A

1. 2.1
2. 3.9
3. 4.2
4. 5.89

Answer: 1. 2.1

2. The number of photons emitted per second from a light source of power 40 W and wavelength 5893 A

1. 3.95 ×10¹¹
2. 1.186 ×10²⁰
3. 3.56 ×10²⁰
4. 3.56×10²⁸

Answer: 2. 1.186 ×10²⁰

Photoelectric effect multiple choice questions 

3. The number of photons emitted per second by a source of light of power 30 W is 10²⁰; the momentum of each photon (in kg – m. s^-1 )

1. 10^-24
2. 10^-25
3. 10^-26
4. 10^-37

Answer: 4. 10^-37

4. Two stationary electrons accelerated with potential differences V₁ and V₂ respectively such that V₁ : V₂ = n . The ratio of their de Broglie Wavelength

1. \(\sqrt{n}\)
2. \(\frac{1}{\sqrt{n}}\)
3. n2
4. \(\frac{1}{n^2}\)

Answer: 2. \(\frac{1}{\sqrt{n}}\)

5. A proton is 1836 times heavier than an electron and has the same charge as that of the electron. For what velocity of the proton will its de Broglie wavelength’ be 4455 A °?

1. 10⁶ m.s^-1
2. 10⁷ m.s^-1
3. 3 × 10⁶ m.s^-1
4. 3 × 10⁷m.s ^-1

Answer: 3.3 × 10⁶ m.s^-1

Question 48. Einstein’s equation for photoelectric effect is Emax = ty-Wo’ where h = Planck’s constant = 6.625 3 × 10^-34 m.s^-1J.s , f = frequency of light incident on metal surface, WQ = work function of metal and E = maximum kinetic energy of the emitted photoelectrons. It is evident that if the frequency is less than a minimum value of f₀ or if the wavelength λ is greater than a maximum value λ₀, the value of E_max would be negative, which is impossible. Thus for a particular metal surface, λ₀ is the threshold frequency f₀ is the threshold wavelength for photoelectric emission to take place.

Again the collector plate is kept at a negative potential concerning the emitter plate, the velocity of the photoelectrons would decrease. The minimum potential for which the velocity of the speediest electron becomes zero is known as the stopping potential, the photoelectric effect stops for a potential lower than this [velocity of light = 3× 10⁸ m.s^-1; the mass of an electron, m = 9.1× 10^-31 kg; charge of an electron, e = 1.6× 10¹⁹ C]

1. The. threshold wavelength of the photoelectric effect for a metal surface is 4600 A. The work function of the metal (in eV) is

1. 2.7
2. 3.45
3. 4.2
4. 6.9

Answer: 1. 2.7

2. Ultraviolet ray of wavelength 1800 A° is incident on the metal surface. The maximum velocity of the emitted photoelectron (in m s_1) is

1. 8.5 × 10⁵
2. 1.2 × 10⁶
3. 1.7 × 10⁶
4. 2.4 × 10⁶

Answer: 2. 1.2 × 10⁶

WBCHSE physics dual nature MCQs 

3. The stopping potential in case of an incident ultraviolet ray of wavelength 1800 A (in V) is

1. 2.7
2. 3.45 V
3. 4.2
4. 6.9

Answer: 3. 4.2

Question 49. The wavelength of matter waves associated with an electron of mass m having kinetic energy E is given by (h is Planck’s constant)

1. \(\frac{2 h}{m E}\)
2. 2mhE
3. \(\frac{2 \sqrt{2 m E}}{h}\)
4. \(\frac{h}{\sqrt{2 m E}}\)

Answer: 4. \(\frac{h}{\sqrt{2 m E}}\)

If p is the momentum, E = \(\frac{p^2}{2 m}\)

Or, p = \(\sqrt{2 m E}\)

Wavelength ofthe associated matter wave = \(\frac{h}{p}=\frac{h}{\sqrt{2 m E}}\)

Physics MCQs on Dual Nature of Matter and Radiation

Question 50. For a monochromatic light incident on a metal surface, the maximum velocity of the emitted photoelectrons is v . Then the stopping potential would be

1. \(\frac{2 m v^2}{e}\)
2. \(\frac{m v^2}{e}\)
3. \(\frac{m v^2}{2 e}\)
4. \(\frac{m v^2}{\sqrt{2} e}\)

Answer: 3. \(\frac{m v^2}{2 e}\)

If V₀ is the stopping potential, then the maximum energy of emitted photoelectrons = eV₀

∴ eV₀ =½mv²

Or, V₀= \(\frac{m v^2}{2 e}\)

Question 51. When green light is incident on a certain metal surface, electrons are emitted but no electrons are emitted with yellow light. If the red light is incident on the same metal surface

1. More energetic electrons will be emitted
2. Less energetic electrons will be emitted
3. The emission of electrons will depend on the intensity of light
4. No electrons will be emitted

Answer: 4. No electrons will be emitted

The energy of photons of red light is less than that of yellow light

Question 52. The wavelength of de Broglie waves associated with a thermal neutron of mass m at absolute temperature T is given by (k is the Boltzmann constant)

1. \(\frac{h}{\sqrt{m k T}}\)
2. \(\frac{h}{\sqrt{2 m k T}}\)
3. \(\frac{h}{\sqrt{3 m k T}}\)
4. \(\frac{h}{2 \sqrt{m k T}}\)

Answer:  3. \(\frac{h}{\sqrt{3 m k T}}\)

E = \(\frac{3}{2}\) kT

Again  E = ½mv²

Or, mv = \(\sqrt{2 m E}=\sqrt{2 m \cdot \frac{3}{2} k T}\)

= \(\sqrt{3 m k T}\)

de Broglie wavelength

λ = \(\frac{h}{m v}=\frac{h}{\sqrt{3 m k T}}\)

WBCHSE physics dual nature MCQs 

Question 53. Find the correct statements about the photoelectric effect.

1. There is no significant time delay between the absorption of suitable radiation and the emission of electrons.
2. Einstein analysis gives a threshold frequency above which no electron can be emitted.
3. The maximum kinetic energy of the emitted photoelectrons is proportional to the frequency of incident radiation.
4. The maximum kinetic energy of electrons does not depend on die intensity of radiation

Answer: 1 And 4

Question 54. The de Broglie wavelength of an electron is the same as that of a 9.50 KeV X-ray photon. The ratio of the energy of the photon to the kinetic energy of the electron is (the energy equivalent of electron mass is 03 MeV)

1. 1:50
2. 1:20
3. 20:1
4. 50:1

Answer: 3. 20:1

The energy of photons,

E = \(\frac{h c}{\lambda}\)

= 50 × 10³ eV

The kinetic energy of the electron

K = ½mv²

= \(\frac{h^2}{2 m \lambda^2}\)

Since v = h/mλ

= \(\frac{E}{K}=\frac{h c}{\lambda} \times \frac{2 m \lambda^2}{h^2}\)

= \(\frac{2 m \cdot c^2}{\left(\frac{h c}{\lambda}\right)}=\frac{2 \times 0.5 \times 10^6}{50 \times 10^3}=\frac{20}{1}\)

E:K = 20:1

Question 55. The work function of metals is in the range of 2 eV to 5 eV. Find which of the following wavelengths of light cannot be used for the photoelectric effect (Consider, Planck’constan = 4 × 10^-15 eV s, the velocity of light
= 3 × 10⁸ m/s )

1. 510 nm
2. 600 nm
3. 400 nm
4. 570 nm

Answer: 2. 600 nm

The energy of the incident photons should not be less than 2 eV

The energy of the photon, E = hv = hc/λ

Hence maximum value of the wavelength

λ_m = hc/E = \(\frac{\left(4 \times 10^{-15}\right) \times\left(3 \times 10^4\right)}{2}\)

= 6 × 10^-7m

= 6 × 10^-9m

=  600 nm

Therefore light of wavelength 600 nm cannot be used for a photoelectric effect

Question 56. Consider two particles of different masses. in which of the following situations the beaver of the two particles will have a smaller de Broglie wavelength?

1. Both have a free fall through the same height
2. Both move with the same kinetic energy
3. Both move with the same linear momentum
4. Both move with the same speed

Answer: 1, 2 And 4

de Broglie wavelength

λ = \(\frac{h}{p}=\frac{h}{m v}\)

= \(\frac{h}{\sqrt{2 m E}}\)

h = Planks constant , m = mass, v= velocity, p= momentum, E = Kinetic energy

I the linear momentum of the two particles are equal their ewavelenght s will also be equal. The velocities of the particles are the same in both cases 1 And 4, hence will be relatively smaller for the heavier particle, Similarly, the E of the two particles i the same for case B, hence will be smaller again for the heavier particle.

WBCHSE physics dual nature MCQs 

Question 57. The potential difference V required for Accelerating an electron to have the de Broglie of 1/λ is wavelength of 1/λ  is

1. 100 V
2. 125 V
3. 150 V
4. 200V

Answer: 3. 150 V

de Broglie wavelength of an electron

λ = \(\frac{h}{\sqrt{2 m q V}}\)

or, V=  \(\frac{h^2}{2 m q \lambda^2}\)

= \(\frac{\left(6.6 \times 10^{-34}\right)^2}{2 \times 9.1 \times 10^{-31} \times 1.6 \times 10^{-19} \times\left(1 \times 10^{-10}\right)^2}\)

= 1.496 × 10² ≈ 150 V

Question 58. The work function of cesium is 2.27 eV. The cut-off voltage which stops the emission of electrons from a cesium cathode irradiated with light of 600 nm wavelength is

1. 0. 5 V
2. – 0.2 V
3. -0.5 V
4. 0.2 V

Answer: None of these

The energy of each photon incident on cesium

E = \(\frac{h c}{\lambda}=\frac{6.6 \times 10^{-34} \times 3 \times 10^8}{600 \times 10^{-9}}\)

= 3.3 × 10^-19 J

= 2.06 eV

Which is less than the work function of cesium (2.27 eV). Hence there will not be any photocurrent

Question 59. The distance between a light source and a photoelectric cell is d. The distance is decreased to \(\) then

1. The emission of electrons per second will be four times
2. The maximum kinetic energy of photoelectrons will be four times
3. Stopping potential will remain the same
4. The emission of electrons per second will be doubled

Answer: 1 And 3

The intensity oflight incident on the photoelectric cell

I ∝ \(\frac{1}{d^2}\)

When the distance between the light source and the photoelectric cell is d.

Again, the number of photoelectrons emitted, n ∝ I

∴ n = \(\frac{1}{d^2}\)

Or = \(\frac{n_2}{n_1}=\left(\frac{d_1}{d_2}\right)^2\)

Or = \(n_2=n_1 \times\left(\frac{d_1}{d_2}\right)^2\)

= \(n_1\left(\frac{d}{\frac{d}{2}}\right)^2\)

= 4n₁

Maximum kinetic energy of photoelectrons,

E_max = eV₀ = hf – W₀

As the frequency (f) of incident light remains the same, the maximum kinetic energy (E_max) and stopping potential (E_max) will remain unchanged.

Question 60. The de Broglie wavelength of an electron is 0.4 ×10^-10m when its kinetic energy is 1.0 keV. Its wavelength will be 1.0  ×10^-10m when its kinetic energy is

1. 0.2 keV
2. 0.8 keV
3. 0.63 keV
4. 0.16 keV

Answer: 4. 0.16 keV

de Broglie wavelength, λ = \(\frac{h}{p}=\frac{h}{\sqrt{2 m E_k}}\)

[p = momentum, E_k =  kinetic energy]

∴ \(\frac{\lambda_1}{\lambda_2}=\sqrt{\frac{\left(E_k\right)_2}{\left(E_k\right)_1}}\)

∴ \(\left(E_k\right)_2=\left(\frac{\lambda_1}{\lambda_2}\right)^2 \times\left(E_k\right)_1\)

= \(\left(\frac{0.4 \times 10^{-10}}{1.0 \times 10^{-10}}\right)^2 \times 1\)

= 0.16 keV

Question 61. When light of frequency ν₁ is incident on a metal with work function W (where hν₁>W), the photocurrent falls to zero at a stopping potential of V₁. If the frequency of light is increased to ν₂, the stopping potential changes to V₂. Therefore the change of an electron is given by.

1. \(\frac{W\left(\nu_2+\nu_1\right)}{\nu_1 V_2+\nu_2 V_1}\)
2. \(\frac{W\left(\nu_2+\nu_1\right)}{\nu_1 V_1+\nu_2 V_2}\)
3. \(\frac{W\left(\nu_2-\nu_1\right)}{\nu_1 V_2-\nu_2 V_1}\)
4. \(\frac{W\left(\nu_2-\nu_1\right)}{\nu_2 V_2-\nu_1 V_1}\)

Answer: 3.\(\frac{W\left(\nu_2-\nu_1\right)}{\nu_1 V_2-\nu_2 V_1}\)

According to Einstein’s photoelectric equation

h = W+eV₁ …………………………(1)

And = W+eV₂ …………………….(2)

Solving equation (1) and (2) we get

e = \(\frac{W\left(\nu_2-\nu_1\right)}{\nu_1 V_2-\nu_2 V_1}\)

Question 62. An electron accelerated through a potential of 10000 V from rest has a de Broglie wavelength A. What should be the accelerating potential so that the wavelength is 4 doubled? a

1. 20000 V
2. 40000 V
3. 5000 V
4. 2500 V

The stored energy in the electron accelerated through a potential of 10000 V

E = hf

Or, 10000 e = \(\) Where e = charge of electron

= de Broglie wavelength of the electron

Or, 5000 e = hc/2

The accelerating potential should be 5000 V so that the wavelength is doubled

WBCHSE physics dual nature MCQs 

Question 63. Radiation of wavelength A is incident on a photocell. The fastest emitted electron has speed v. If the wavelength is changed to \(\frac{3}{4}\), the speed of the fastest emitted electron will be

1. \(>v\left(\frac{4}{3}\right)^{1 / 2}\)
2. \(<v\left(\frac{4}{3}\right)^{1 / 2}\)
3. \(=v\left(\frac{4}{3}\right)^{1 / 2}\)
4. \(=v\left(\frac{3}{4}\right)^{1 / 2}\)

Answer: 1. \(>v\left(\frac{4}{3}\right)^{1 / 2}\)

In the first case \(\frac{1}{2} m v^2=\frac{h c}{\lambda}-W_0\) ………. (1)

In the second case,

⇒ \(\frac{1}{2} m v^2=\frac{h c}{\lambda}-W_0\)

[where = speed of fastest emitted electron when wavelength is \(\frac{3}{4}\)

Or, \(\frac{1}{2} m v_1^2=\frac{1}{2} m v^2+\frac{h c}{3 \lambda}\)

Using equation (1)

Or, \(\sqrt{v^2+\frac{2 h c}{3 \lambda m}}=\sqrt{v^2+\frac{2}{3 m}\left(\frac{1}{2} m v^2+W_0\right)}\)

= \(\sqrt{\frac{4 v^2}{3}+\frac{2 W_0}{3 m}}\)

= \(v_1>\sqrt{\frac{4}{3} v^2} \quad \text { or, } v_1>\sqrt{\frac{4}{3}} v\)

Question 64. A particle A of mass m and initial velocity v collides with a particle B of mass y which is at rest. The collision is head-on and elastic. The ratio of the de Broglie wavelengths AA to AB after collision is

1. \(\frac{\lambda_A}{\lambda_B}=\frac{1}{3}\)
2. \(\frac{\lambda_A}{\lambda_B}\) = 2
3. \(\frac{\lambda_A}{\lambda_B}=\frac{2}{3}\)
4. \(\frac{\lambda_A}{\lambda_B}=\frac{1}{2}\)

Answer: 2. \(\frac{\lambda_A}{\lambda_B}\) = 2

Let after collision the velocities of A and B respectively vA and vB.

According to the law of conservation of momentum

mv = \(m v_A+\frac{m}{2} v_B\)

According to the law of conservation of relative velocity

v = v_B – v_A

Solving (1) and (2) we get, vA = \(\) and \(\frac{4 u}{3}\)

= \(\frac{\lambda_A}{\lambda_B}=\frac{\frac{h}{p_A}}{\frac{h}{p_B}}=\frac{p_B}{p_A}\)

= \(\frac{\frac{m}{2} v_B}{m v_A}=\frac{\frac{4 v}{3 \times 2}}{\frac{v}{3}}\)

= 2

Question 65. When the energy of the incident radiation is increased by 20%, the kinetic energy of the photoelectrons emitted from a metal surface increases from 0.5eV to 0.8eV. The work function of the metal is

1. 0. 65 eV
2. 1.0 eV
3. 1.3 eV
4. 1.5 eV

Answer: 2. 1.0 eV

According to Einstein’s equation,

E = hf – W₀ or,hf = E + W₀

For the first case, hf₁ = E₁ + W₀

For the second case, hf₂ = E₂ + W₀

⇒ \(\frac{h f_2}{h f_1}=\frac{E_2+W_0}{E_1+W_0} \text { or, } \frac{120}{100}=\frac{0.8+W_0}{0.5+W_0}\)

Or, 3+ 6W₀ = 4+5 W₀

9W₀ = 9W₀

W₀ =  \(\frac{9}{9}\)

W₀ =  1eV

WBCHSE physics dual nature MCQs 

Question 66. If the kinetic energy of the particle is increased to 16 times its previous value, the percentage change in the de Broglie wavelength of the particle is

1. 25%
2. 75%
3. 60%
4. 50%

Answer: 2. 75%

Wavelength = \(\)

= \(\lambda=\frac{h}{p}=\frac{h}{\sqrt{2 m E}} \quad \text { or, } \lambda \propto \frac{1}{\sqrt{E}}\)

= \(\frac{\lambda_2}{\lambda_1}=\sqrt{\frac{E_1}{E_2}}=\frac{1}{4}\)

Or, = 4λ₂

∴ Percentage change in wavelength

= \(\frac{\lambda_1-\lambda_2}{\lambda_1} \times 100=\frac{4 \lambda_2-\lambda_2}{4 \lambda_2} \times 100\)

= 75

Question 67. Radiation of energy E falls normally on a perfectly reflecting surface. The momentum transferred to the surface is (c = velocity of light)

1. E/c
2. 2E/c
3. 2E/c²
4. E/c²

Answer: 2. 2E/c

Transfer of momentum for an incident photon of energy hf is hf/c. Hence, the transfer of momentum for the incidence of energy

E is E/c

Again, the transfer of momentum due to the reaction for total reflection of radiation is E/c

∴ The total momentum transferred = 2E/c

Question 68. A certain metallic surface is illuminated with monochromatic light of wavelength, A. The stopping potential for photoelectric current for this light is 3 V₀. If the same surface is illuminated with light of wavelength 2λ, the stopping potential is V₀ The threshold wavelength for this surface for the photoelectric effect is

1. 6λ
2. 4λ
3. λ/4
4. λ/6

Answer: 2. 4λ

eV₀ = h(f-f₀)

Or, eV₀ = \(\left(\frac{c}{\lambda}-f_0\right)\)

f₀ = Threshold frequency

In the first case

e. 3V₀ = \(\left(\frac{c}{\lambda}-f_0\right)\)………………….. (1)

In the second case

= \(\left(\frac{c}{2 \lambda}-f_0\right)\)……………………… (2)

Dividing equation (1 ) by equation (2)

3 =  \(=\frac{\frac{c}{\lambda}-f_0}{\frac{c}{2 \lambda}-f_0}\)

Or, \(\frac{c}{\lambda}-f_0=\frac{3}{2} \cdot \frac{c}{\lambda}-3 f_0\)

Or,  2f₀ = \(\frac{1}{2} \frac{c}{\lambda} \quad \text { or, } f_0=\frac{c}{4 \lambda}\)

∴ Threshold wavelength, λ₀ = c/f₀ = 4λ

Practice Questions on Wave-Particle Duality WBCHSE

Question 69. Which of the following figures represents the variation of the particle momentum and the associated de Broglie wavelength?

Answer: 2.

de Broglie wavelength, \(\)

p = Momentum

So, pλ = h = constant

The λ – p graph will be a rectangular hyperbola.

Question 70. When a metallic surface is illuminated with radiation of wavelength λ, the stopping potential is V. If the same surface Is illuminated with radiation of wavelength 2 λ, the stopping potential is V/4 The threshold wavelength for the metallic surface is

1. 5λ
2. 5/2λ
3. 3λ
4. 4λ

Answer: 3. 3λ

In the first eV = \(\frac{h c}{\lambda}-W_0\)  ……………………………….(1)

In the second case = \(e \frac{V}{4}=\frac{h c}{2 \lambda}-W_0\) ………………………………(2)

Subtracting equation (2) × 4 from equation (1 ), we get,

0 = \(-\frac{h c}{\lambda}+3 W_0\)

Or, \(W_0=\frac{h c}{3 \lambda}\)

Threshold wavelength, λ₀ = 3λ

Question 71. An electron of mass m and a photon have the same energy E. The ratio of de Broglie wavelengths associated with them is

1. \(\left(\frac{E}{2 m}\right)^{1 / 2}\)
2. \(c(2 m E)^{1 / 2}\)
3. \(\frac{1}{c}\left(\frac{2 m}{E}\right)^{1 / 2}\)
4. \(\frac{1}{c}\left(\frac{E}{2 m}\right)^{1 / 2}\)

Answer: 4.

de Broglie wavelength of electron = \(\lambda_e=\frac{h}{p}=\frac{h}{\sqrt{2 m E}}\)

de Broglie wavelength of photon = \(\lambda_p=\frac{h c}{E}\)

∴ \(\frac{\lambda_e}{\lambda_p}=\frac{h}{\sqrt{2 m E}} \times\)\(\frac{E}{h c}\)

= \(\frac{1}{c} \sqrt{\frac{E}{2 m}}\)

Question 72. If the mass of a neutron is 1.7 × 10^-27 kg, then the de (*ÿ50 Broglie wavelength of neutron of energy 3 eV is h = 6.6 × 10^-34 J.s)

1. 1.4 × 10^-1111m
2. 1.6 × 10^-10 m
3. 1.65 × 10^-11m
4. 1.4 × 10^-10m

Answer: 4. 1.4 × 10^-10m

= \(\frac{h}{p}=\frac{h}{\sqrt{2 m E}}\)

= \(\frac{6.6 \times 10^{-34}}{\sqrt{\left[2 \times\left(1.7 \times 10^{-27}\right) \times\left(3 \times 1.6 \times 10^{-19}\right)\right]}}\)

1.634 × 10^-10m

Question 73. In an experiment of photoelectric effect, the stopping potential was measured to be V₁  and V₂ with incident light of wavelength λ and λ/2 respectively. The relation V₁ and V₂ is

1. V₂ >2V₁
2. V₂ >V₁
3. V₁ <V₂< 2V₁
4. V₂ = 2V₁

Answer: 1. V₂ >2V₁ 

ev = hf – W₀

Or \(\frac{h c}{\lambda}=e V+W\)

For these two cases \(\frac{h c}{\lambda}=e V_1+W\) eV1+ W

And \(\frac{h c}{\lambda / 2}=e V_2+W\) hc/ = ev₁+W₀

Hence, \(\frac{2 h c}{\lambda}=e V_2+W\)

∴ eV₁ + W = 2eV₁ + 2W

V₂ = 2V₁ + \(\frac{W}{e}\)

So, V₂ > 2V₁

Question 74. An electron of mass m with an initial velocity \(\vec{V}=V_0 \hat{i}\), field (V>0) enters an electric \(\vec{E}=E_0 \hat{i}\) E₀ = constant > 0) at t = 0. If λ₀ is its de Broglie de Broglie wavelength of electron

1. λ₀t
2. \(\lambda_0\left(1+\frac{e E_0}{m V_0} t\right)\)
3. \(\frac{\lambda_0}{\left(1+\frac{e E_0}{m V_0} t\right)}\)
4. λ₀ 

Answer: 3. \(\frac{\lambda_0}{\left(1+\frac{e E_0}{m V_0} t\right)}\)

Initial de Broglie wavelength of the electron,

⇒ \(\lambda_0=\frac{h}{m V_0}\)

Velocity ofthe electron after time t

V = u+ at = \(V_0+\frac{F}{m} t=V_0+\frac{e E_0}{m} t\)

= \(V_0\left[1+\frac{e E_0}{m V_0} t\right]\)

= \(\frac{h}{m V}=\frac{h}{m V_0\left[1+\frac{e E_0}{m V_0} t\right]}\)

= \(\frac{\lambda_0}{\left[1+\frac{e E_0}{m V_0} t\right]}\)

Study Guide for Dual Nature of Matter WBCHSE

Question 75. When the light of frequency 2ν₀ (where l/Q is threshold frequency) is incident on a metal plate, the maximum velocity of electrons emitted is v₁. When the frequency of the incident radiation is increased to 5ν₀, the maximum velocity ofelectrons emitted from the same plate is v₂. The ratio of v₁ to v₂ is

1. 4: 1
2. 1:4
3. 1:2
4. 2:1

Answer: 3. 1:2

E = \(W_0+\frac{1}{2} m v^2\)

First case \(h \nu_0+\frac{1}{2} m v_1^2\)

Or, \(h \nu_0=\frac{1}{2} m \nu_1^2\) …………….. (1)

For the second case \(=h \nu_0+\frac{1}{2} m \nu_2^2\)

Or, \(-4 h \nu_0=\frac{1}{2} m v_2^2\)

Now dividing (1) by (2), we have

⇒ \(\frac{\nu_1^2}{\nu_2^2}=\frac{h \nu_0}{4 h \nu_0}=\frac{1}{4}\)

Or, \(\frac{v_1}{v_2}\) = 1:2

Question 41. The wavelength of de Broglie waves associated with a thermal neutron of mass m at absolute temperature T is given by (k is the Boltzmann constant)
Answer:

1. \(\frac{h}{\sqrt{m k T}}\)
2. \(\frac{h}{\sqrt{2 m k T}}\)
3. \(\frac{h}{\sqrt{3 m k T}}\)
4. \(\frac{h}{2 \sqrt{m k T}}\)

Answer: 3. \(\frac{h}{\sqrt{3 m k T}}\)

E \(\frac{3}{2}\)

Again E = ½ mv²

Or, mv = \(\sqrt{2 m E}\)

= \(\sqrt{2 m \cdot \frac{3}{2} k T}=\sqrt{3 m k T}\)

de Broglie wavelength \(\frac{h}{m v}=\frac{h}{\sqrt{3 m k T}}\)

Unit 7 Dual Nature Of Matter And Radiation Synopsis

When the surface of a substance is irradiated by light of a suitable wavelength, electrons are emitted from that surface. This phenomenon is known as the photoelectric effect or photo¬ electric emission.

1. The electrons emitted in the photoelectric effect are called
photoelectrons.

2. With the help of suitable arrangement a stream of unidirectional photoelectrons can be obtained and the electric current thus produced is called photoelectric current.

3. The minimum amount of energy required to remove an electron from the surface of a particular substance, is called the work function of that substance.

4. The minimum negative potential of the anode concerning the photocathode, for which photoelectric current becomes zero is called the stopping potential.

5. The value of stopping potential depends on

• The nature of the metallic surface of the photocathode and
• The wavelength or frequency of the incident light.

6. The minimum frequency of the incident radiation required to emit photoelectrons from the surface of a substance, is called the threshold frequency for that substance.

7. Photoelectric emission is an instantaneous process.

8. The cell made based on the photoelectric effect where light energy is converted into electrical energy, is called a photoelectric cell.

9. According to de Broglie’s hypothesis, matter also behaves as waves.

10. The photoelectric effect cannot be explained in terms of the wave theory oflight. Einstein first introduced the concept of photon particles by using Planck’s quantum theory. Elec¬ tromagnetic radiation consists of a stream of particles. These particles are called photons

Unit of Planck’s constant (h) :

Unit of h in SI = J .s

Unit of h in CGS system = org-s

Unit of h In eV = eV.s

11. Einstein’s photoelectric equation is based on the quantum theory of radiation. This equation correctly explains the following observations in the photoelectric effect.

1. Maximum kinetic energy of photoelectrons,
2. Threshold frequency,
3. Photoelectric emission is Instantaneous,
4. Dependence of photoelectric current on Intensity of incident light.

12. Radiation sometimes behaves like waves and sometimes like a stream of particles. Thus, wave theory and particle theory are not contradictory but complementary to each other.

13. According to de Broglie’s hypothesis, the concept of matter waves is only important in the case of particles of atomic dimensions.

14. In 1927, two American scientists Davisson and Germer first experimentally, demonstrated diffraction of electrons to prove the existence of matter waves.

15. The matter wave cannot be represented by a pure sinusoidal wave. Matter waves can be represented by duly formed wave groups or wave packets.

16. Matter wave is neither a type of elastic wave nor a magnetic wave.

17. At any instant, a moving particle is located at a specific point but at that instant matter wave associated with that particle extends over some space. This is an inherent property of matter-wave

18. The relation between the kinetic energy of photoelectrons and the stopping potential is,

½ mv²_max = eV₀

Or, \(\sqrt{\frac{2 e V_0}{m}}\)

m – mass of electron having charge e, v_max = maximum initial velocity of photoelectron and V₀ =stoppingpotential

4 If the threshold frequency is f₀,

Threshold wavelength,

λ₀ = \(\frac{c}{f_0}\) = c( velocity of light)

Amount of energy carried by a photon, E = hf.

Here, f = frequency of the radiation, h = Planck’s constant

19. According to the theory of relativity, if the rest mass of a particle is m₀ and its momentum is p, the energy of the particle

E = \(\sqrt{p^2 c^2+m_0^2 c^4}\)

In the case of a photon m₀ = 0

Hence E = pc

∴ p = \(\frac{E}{c}=\frac{h f}{c}\)

20. The relation between the wavelength of radiation and photon energy is E = 12400/ λ (inÅ)

21. Einstein’s photoelectric equation can be written as,

E_max  = hf-W₀

Or, ½mv² =  hf-W₀

Or, ev₀ = hf- W₀

22. When a radiation of frequency f is incident on a metal of threshold frequency f₀(f > f₀)’, ten the maximum kinetic energy of emitted photoelectrons

E_max= hf-hf₀= h(f-f₀)

Since (W₀= hf₀)

= hc \(\left(\frac{1}{\lambda}-\frac{1}{\lambda_0}\right)\)

Where λ and λ₀ are the wavelength of the incident light and threshold wavelength for the metal surface, respectively.

This equation is another form of Einstein’s photoelectric equation. Important information

h = 6.625 ×10^-12 erg. s , c = 3 ×10¹⁰ cm .s^-1

1 eV = 1.6 × 10^-12 erg

22. de Broglie wavelength, λ = \(\frac{h}{p}\)

Where, p = mv – momentum of the particle

23. If an electron is accelerated by V volt, then the de Broglie wavelength associated with the moving electron

λ =   \(\frac{12.27}{\sqrt{V}}\) Å

24. de Broglie wavelength of any molecule (mass m ) of gas at temperature TK,

λ =   \(\frac{h}{\sqrt{3 m k T}}\)

(k = Boltzmann constant)

25. Number of photons of wavelength X emitted from a lamp of power P in time t is

n = \(\frac{p t \lambda}{h c}\)

26. Photoclectrons are easily produced from the surface of metals that have a low work function. For example, cesium.

WBCHSE Class 12 Physics MCQs

Diffraction And Polarisation Of Light Multiple Choice Questions

Question 1. A light beam is i incident from ait to a glass slab at Brewster angle, A polaroid placed in the path of the refracted beam is being rotated about an axis passing through the center and perpendicular to the plane of the polaroid

1. For a particular orientation, there shall be darkness as observed through the polaroid
2. The intensity of light as seen through the polaroid shall be independent of the rotation
3. The intensity of light as seen through the polaroid shall go through a maximum but not zero for two orientations of the polaroid
4. The intensity of light as seen through the polaroid shall go through a minimum but not zero for two orientations of the polaroid

Answer: 3

Question 2. Consider the diffraction pattern of a small pinhole. As the size of the hole Is Increased the changes that occur In the pattern

1. The size decreases
2. The intensity increases
3. The size Increases
4.  The intensity decreases

Answer: 1 And 2

Question 3. Geometrical optics is valid when the dimensions of the aperture are

1. Of the same order as the wavelength of light
2. Much smaller than the wavelength of light
3. Most the larger than the wavelength of light
4. Of the order of 1 A

Answer: 3. Much of the larger than the wavelength of light

Read and Learn More Class 12 Physics Multiple Choice Questions

Question 4. Light appears to propagate in straight lines because

1. It is reflected by the upper part of the atmosphere
2. It is not absorbed in the atmosphere
3. Its speed is very large
4. Its wavelength is very small

Answer: 4. Its wavelength is very small

Question 5. Which of the following does not support the corpuscular nature of light?

1. Photoelectric effect
2. Compton effect
3. Diffraction
4. Blackbody radiation

Answer: 2. Compton effect

WBBSE Class 12 Diffraction MCQs

Question 6. Maximum diffraction takes place in the case of

1. Ultraviolet rays
2. Radio waves
3. γ-rays
4. Infrared waves

Answer: 4. Infrared waves

Question 7. Among the following conditions which one is an essential condition for Fresnel diffraction?

1. The source and screen will remain at infinite distance from the slit
2. Either source or screen will remain at infinite distance from the slit
3. Neither source nor screen will remain at infinite distance from the slit
4. None of the above

Answer: 3. Neither source nor screen will remain at infinite distance from the slit

Question 8. In single slit diffraction of light of wavelength λ, the angular width of the central maximum (considering slit width d) will be

1. \(\frac{d}{\lambda}\)
2. \(\frac{\lambda}{d}\)
3. \(\frac{2 \lambda}{d}\)
4. \(\frac{2 d}{\lambda}\)

Answer: 3. \(\frac{2 \lambda}{d}\)

Question 9. In the diffraction of light of wavelength λ at a single slit of width a, the angle θ between the central maximum and first minimum on either side is

1. \(\frac{d}{\lambda}\)
2. \(\frac{\lambda}{2 a}\)
3. \(\frac{\lambda}{4 a}\)
4. \(\frac{\pi}{2}\)

Answer: 1. \(\frac{d}{\lambda}\)

Question 10. The width of single-slit diffraction fringes varies

1. Directly with the distance between the slit and the fringes
2. Inversely with the wavelength of light
3. Directly with the width of the slit
4. None of above

Answer: 2. Inversely with the wavelength of light

Conceptual Questions on Wave Optics

Question 11. If the width of the slit is increased, what changes would be there in the linear width of the central maximum?

1. Increases
2. Decreases
3. Remains unchanged
4. None of the above

Answer: 2. Decreases

Question 12. In a single slit (width = a) diffraction of light wavelength λ, fringes are produced with a diffraction angle θ. The condition of formation of the first minimum is

1. λsin θ = a
2. a cos θ  = λ
3. a sin θ = λλcos θ= a

Answer: 3. a sin θ = λ

Question 13. In a single slit experiment, the angular width of the first minima obtained by using the light of wavelength G980 A is 2°. The width of the slit is

1. 2 × 10^-5mm
2. 0.02 mm
3. 0.2mm
4. 2 mm

Answer: 2. 0.02mm

Question 14. In a single slit experiment, the angular width of the first minima obtained by using the light of wavelength 6980 A° is 2°. The width of the slit is

1. 2.4 mm
2. 4.8 mm
3. 9.6 mm
4. None of these

Answer: 4. None of these

WBCHSE class 12 physics MCQs 

Question 15. The width of the diffraction band is

1. Inversely proportional to the wavelength of the incident light
2. Inversely proportional to the size of the source which illuminates the slit.
3. Directly proportional to the distance between the slit and the screen
4. Directly proportional to the width of the slit

Answer: 3. Directly proportional to the distance between the slit and the screen

Question 16. In a single slit Fraunhofer diffraction experiment, the wavelength of the light used is 400 nm and the first minimum is obtained for a diffraction angle of 30°. The magnitude of 8 for first secondary maximum is

1. \(\sin ^{-1}\left(\frac{2}{3}\right)\)
2. \(\sin ^{-1}\left(\frac{3}{4}\right)\)
3. \(\sin ^{-1}\left(\frac{1}{4}\right)\)
4. \(\sin ^{-1}\left(\frac{2}{3}\right)\)

Answer: 2. \(\sin ^{-1}\left(\frac{3}{4}\right)\)

Question 17. In a single slit experiment, a beam of parallel monochro¬ matic light falls on a slit perpendicularly. A diffraction fringe is produced on a screen kept perpendicular to the direction of propagation of light. What would be the phase difference of the two rays coming from two ends of the slit at the first minimum of fringe?

1. 0
2. \(\frac{\pi}{2}\)
3. π
4. 2π

Answer: 4.

Question 18. In a single-slit experiment, the tire width of the slit is reduced. The linear width of the central maximum

1. Will Increase but the intensity will be reduced
2. Will decrease but Intensity will be increased more
3. Will increase but the Intensity will be Increased more
4. Will decrease but the intensity will be reduced

Answer: 1. The linear width of the central maximum

Question 19. In a single-slit experiment, the width of the slit is reduced to half. To keep the width of the central maximum the same, what is to be done of the following?

1. The distance between the slit and the screen is to be reduced to half
2. The distance between the slit and the screen is to be reduced to one-fourth
3. The distance between the slit and the screen is to be doubled
4. Nothing to be done, as the width of central maximum does not depend upon the width of the slit

Answer: 1. The distance between the slit and the screen is to be reduced to half

Practice Questions on Young’s Double Slit Experiment

Question 20. A narrow slit of 2 mm width is illuminated by a monochromatic light of wavelength 500 nm. What would be the intermediate distance between two first minima on either side of a screen kept away?

1. 5 mm
2. 0.5 mm
3. 1mm
4. 10mm

Answer: 2. 0.5 mm

Question 21. The resolving power of a microscope is

1. Inversely proportional to the numerical aperture of the lens of the objective
2. Directly proportional to the wavelength of the light
3. Directly proportional to the square of the wavelength of the light
4. Directly proportional to the numerical aperture of the lens of the objective

Answer: 4. Directly proportional to the numerical aperture of the lens of the objective

Question 22. For the minimum angular distance between two stars, which a telescope can analyze, which of the following statements is true (here aperture means the diameter of the objective of the telescope)?

1. Angular distance decreases with the increase in the aperture of the telescope
2. Angular distance does not depend on the aperture of the telescope.
3. Angular distance increases linearly with the aperture of the telescope.
4. Angular distance increases with the second power of the aperture of the telescope

Answer:  1. Angular distance decreases with the increase in the aperture of the telescope

Question 23. An unpolarised light of intensity I_o falls on a pair of Nicol prism. The angle between the two prisms is 60°. the intensity of the light emitted from the prism will be

1. I_o
2. \(\frac{I_0}{2}\)
3. \(\frac{I_0}{4}\)
4. \(\frac{I_0}{8}\)

Answer: 4. \(\frac{I_0}{8}\)

Diffraction and polarization class 12 MCQs 

Question 24. A polaroid is inclined at an angle of 45° with an incident light intensity of 70°. After polarization, the intensity of the light emitted from the polaroid is

1. I_o
2. \(\frac{I_0}{2}\)
3. \(\frac{I_0}{4}\)
4. 0

Answer: 2. \(\frac{I_0}{2}\)

Question 25. Amplitude of unpolarised light on a polariser is a. Amplitude of polarised light which passes through the polariser is

1. \(\frac{a}{2}\)
2. \(\frac{a}{\sqrt{2}}\)
3. \(\frac{\sqrt{3} a}{2}\)
4. \(\frac{3 a}{4}\)

Answer: 2. \(\frac{a}{\sqrt{2}}\)

Question 26. When a light ray in the air is incident on glass at 57°, the reflected rays are completely polarised. The same ray is incident on the water at an angle 6, also the reflected ray is completely polarised. Then

1. θ >57°
2. θ <57°
3. θ = 57°
4. θ = 90°

Answer: 2. θ <57°

Question 27. The Brewster’s law polarisation of fight is

1. μ sin i_p= 1
2. μ cos i_p = 1
3. μ tan i_p = 1
4. μ cot i_p = 1

Answer: 4. μ cot i_p = 1

Question 28. From Brewster’s law, it follows that the polarising angle depends on

1. Wavelength of fight
2. Frequency
3. Plane of polarisation
4. Plane of vibration

Answer: 1. Wavelength of fight

Question 29. The tangent of the polarising angle is equal to the refractive indices of the reflecting medium. This is called

1. Brewster’s law
2. Law of Malus
3. Braggs law
4. Grimaldi’s law

Answer: 1. Brewster’s law

Diffraction and polarization class 12 MCQs 

Question 30. A ray of light is incident on a glass. The reflected ray gets totally polarized. The magnitude of the incidence angle is (refractive index of glass =p)

1. \(\sin ^{-1}(\mu)\)
2. \(\sin ^{-1}\left(\frac{1}{\mu}\right)\)
3. \(\tan ^{-1}\left(\frac{1}{\mu}\right)\)
4. \(\tan ^{-1}(\mu)\)

Answer: 4. \(\tan ^{-1}(\mu)\)

Question 31. The angle of polarization for a medium is 60°. Critical angle

1. \(\sin ^{-1} \sqrt{3}\)
2. \(\tan ^{-1} \sqrt{3}\)
3. \(\cos ^{-1} \sqrt{3}\)
4. \(\sin ^{-1} \sqrt{\frac{1}{3}}\)

Answer: 4. \(\sin ^{-1} \sqrt{\frac{1}{3}}\)

Question 32. An unpolarized ray of light is an Incident on the water surface The incident angle for which the reflected and refracted rays become perpendicular to each other is  \(\mu_w=\frac{4}{3}\)–

1. \(\sin ^{-1} \frac{4}{3}\)
2. \(\tan ^{-1} \frac{3}{4}\)
3. \(\tan ^{-1} \frac{4}{3}\)
4. \(\sin ^{-1} \frac{1}{3}\)

Answer: 3. \(\tan ^{-1} \frac{4}{3}\)

Class 11 Physics	Class 12 Maths	Class 11 Chemistry
NEET Foundation	Class 12 Physics	NEET Physics

Question 33. The critical angle of a medium is sin-1(0.6). The polarising angle for that medium is

1. tan^-1(1.5)
2. sin^-1(0.8)
3. tan^-1(1-6667)
4. tan^-1 (0.6667)

Answer: 3. tan^-1(1-6667)

Question 34. Brewster’s angle at the glass-air interface is 54.74°. If a ray of light is incident on glass from the air at an angle of 45°,  then the magnitude of the refracted angle will be (given tan54.74° = \(\sqrt{2}\))

1. 60°
2. 30°
3. 25°
4. 54.74°

Diffraction and polarization class 12 MCQs 

Question 35. A bird sitting on the top of a light post, which is at the side of the lake. It observed stop should of at height be completely posted, the angle which of polarised theis at indent light with the plane of the surface of the water? (Refractive index of water = 1.33 )

1. 30°
2. 35°
3. 37°
4. 38°

Answer: 3. 37°

Hint: Here, μ = tan i_p or, tan i_p = 1.33

i_p = tan^-1(1.33) = 53°

The angle between the incident ray and the surface of the water,

θ = 90°- i_p = 90° -53° = 37°

Important Definitions in Diffraction and Polarisation

Question 36.  If μ_O and  μ_E  are the refractive indices of the crystal for O-ray and E-ray respectively, then which of the following relations is correct for a negative crystal

1.  μ_E= μ_O
2. μ_E>μ_O
3. μ_E<μ_O
4. μ_E≥μ_O

Answer: 3.μ_E<μ_O

Question 37. In double refraction we get two refracted rays i.e., O-ray and E-ray. Which one of the following statements is correct?

1. Only O-ray is polarised
2. Only E-ray is polarised
3. Both O-ray and E-ray are polarised
4. None of the O-ray and E-ray is polarised

Answer: 3. Both O-ray and E-ray are polarised

Question 38. When a polaroid, is placed In the path of Unlit, I* rotated, the Intensity of light appears to vary but never reduce* to two. The Light Is

1. Unpolarlsed
2. Plane polarised
3. Partially plane polarised
4. No conclusion can be drawn

Answer: 3. Partially plane polarised

Class 12 physics diffraction questions 

Question 39. When a polaroid, placed in the path of light, Is rotated, there is no change intensity of light. The Incident light –

1. Totally polarised
2. partially plane polarised
3. Unpolarised
4. None of above

Answer: 3. Unpolarised

Question 40. Which of the following undergoes larger diffraction?

1. Ultraviolet light
2. Radio waves
3. γ-rays
4. Sound waves.

Answer: 3 And 4

Question 41. The width of single-slit diffraction fringes varies

1. Directly as the distance between the slit and the screen
2. Directly as the width of the slit
3. Inversely as the width of the slit
4. Inversely as the wavelength of light

Answer: 1 And 3

Question 42. Diffraction patterns can be observed with

1. Two narrow sifts
2. A large number of narrow slits
3. One narrow slit
4. One wide slit

Answer: 1, 2, And 3

Class 12 physics diffraction questions 

Question 43. Interference differs from diffraction in that

1. It cannot be observed with white light
2. Unlike diffraction, the interference fringes are of varying intensity
3. Interference minima are perfectly dark and those of diffraction may not be dark
4. Interference fringes may or may not be of the same width but diffraction fringes are never of the same width

Answer: 3 And 4

Question 44. Ordinary light falling at a polarising angle on a glass slab placed in ai r is partly reflected in air and partly refracted in the slab. Then:

1. The reflected light is completely polarised
2. The reflected light is partially polarised
3. The refracted light is completely polarised
4. The refracted light is partially polarised

Answer: 1 And 4

Question 45. A slit of width a is illuminated by red light of wavelength 650 nm. The first diffraction minimum is observed at an angle θ₁ = 5.2° from the direction of the incident beam

1. The width of the slit and in mm

1. 5.12
2. 7.17
3. 3.21
4. 4.25

Answer: 2 . 7.17

2. The angle θ₂ at which the second minimum Is observed is

1. 10.4
2. 8.6
3. 7.5
4. 9.5

Answer: 1. 10.4

Question  46. Light of wavelength 0500 A passes through a slit 0.1 cm wide and forms a diffraction pattern on a screen 1.8 m away.

1. The width of the central maximum in mm is

1. 2.5
2. 1.32
3. 2.34
4. 1.72

Answer: 3. 2.34

2. The width of the central maximum in mm when the apparatus is immersed in water of refractive index is ____

1. 2.12
2. 1.53
3. 2.54
4. 1.755

Answer: 4.1.755

Class 12 physics diffraction questions 

Question 47. If a light of wavelength λ falling on a single slit of width diffracts an angle θ, the conditions of first minima will be 

1. λ sin θ = a
2. a cos θ= λ
3. a sin θ= λ
4. λ cos θ= a

Answer: 3. a sin θ= λ

In this case, the condition of the first minima will be a sin θ= λ

Question 48.  Which one does not change in the polarisation of light?

1. Intensity
2. Phase
3. Frequency
4. None of these

Answer: 3. Frequency

Question 49. Find the right condition(s) for Fraunhofer diffraction due to a single slit.

1. The source is at an infinite distance and the incident beam has converged at the slit
2. The source is near to the slit and the incident beam is parallel.
3. The source is at infinity and the incident beam is parallel.
4. The source is converged at the slit.

Answer:  2 And 3

The beam of light incident on the slit has to be parallel—here the incident wavefront would become parallel wavefront f.’ or this the source could be kept at infinity or the source could be placed near the slit and the parallel beam of light could be made to fall on the slit by using the proper lens.

Question 50. Two booms, A and R, of plane-polarized light with mutually perpendicular planes of polarization are seen through a polaroid. From the position when beam A has maximum Intensity (and beam B has zero intensity), a rotation of polaroid through 30° makes the two beams appear equally bright. If the initial intensities of the two beams arc IA and IB respectively, then l_A/ I_B equals

1. \(\frac{1}{3}\)
2. 3
3. \(\frac{3}{2}\)
4. 1

Answer: 1. \(\frac{1}{3}\)

According to Malus’ law,  I = I_ocos ² θ

I_A’ = I_A cos ² θ₁

= I_A’ cos ² 30°

= \(\frac{3}{4}\) I_A

Since  θ₁= 30°

Again ,I_B’ = I_B cos ² θ₂

= I_B cos² 60°

⇒ \(\frac{1}{4}\) I_B

Since θ₂= 90 °- θ₁= 60°

According to question I_A’ = I_B’

Or, \(\frac{1}{4}\)I_A = \(\frac{1}{4}\)I_B

Or, \(\frac{I_A}{I_B}\)IB = \(\frac{1}{3}\)

Class 12 physics diffraction questions 

Question 51. Assuming the human pupil to have a radius of 0.25 cm comfortable viewing distance of 25 cm, the minimum separation between two objects that the human eye can resolve at 500 nm wavelength Is

1. 1 μm
2. 30 μm
3. 100 μm
4. 300 μm

Answer: 2. 30 μm

Diameter, d = 0.25 × 2 = 0.5 cm

= 0.005 m

D =  25 cm =  0.25 m

λ = 500 nm = 5 ×10^-7m

Angular resolution, A = \(\Delta \theta=\frac{1.22 \lambda}{d}\)

If the required minimum separation is x

= \(\Delta \theta=\frac{\Delta x}{D}\)

= \(\Delta x=D \Delta \theta=0.25 \times \frac{1.22 \times\left(5 \times 10^{-7}\right)}{0.005}\)

= 30.5 ×10^-6m = 30.5μm

= 30.5μm ≈ = 30 μm

Question 52. The Box of n pinhole camera of length L₁ has a hole of radius a. It is assumed that when the hole Is Illuminated by a parallel beam of light of wavelength A, the spread of the spot (obtained on the opposite wall of the camera) Is the sum of Its geometrical spread and the spread due to diffraction. The spot would then have Its minimum size (say when

1. a = \(\frac{\lambda^2}{L} \text { and } b_{\min }=\frac{2 \lambda^2}{L}\)
2. a = \(\sqrt{\lambda L} \text { and } b_{\min }=\frac{2 \lambda^2}{L}\)
3. a = \(\sqrt{\lambda L} \text { and } b_{\min }=\sqrt{4 \lambda L}\)
4. a = \(\frac{\lambda^2}{L} \text { and } b_{\min }=\sqrt{4 \lambda L}\)

Answer: 3. a= \(\sqrt{\lambda L} \text { and } b_{\min }=\sqrt{4 \lambda L}\)

sin θ = \(\frac{\lambda}{a} \)

tan θ  = \(\frac{C E}{A E}\)

Since θ  is very small, so tan θ ≈ sin θ

∴ sin = \(\frac{C E}{A E}\)

Or, = \(\frac{\lambda}{a}=\frac{C E}{L}\)

Or, CE = \(\frac{L \lambda}{a}\)

Hence diameter of the spot,

B = EF + CE + FD = 2a +\(\frac{2 L \lambda}{a}\)

The diameter of the spot will be minimal when

⇒ \(\frac{d B}{d a}\) = 0

or, 2 – \(\frac{2 L \lambda}{a}\) = 0

Or, \(\frac{L \lambda}{a^2}\) = 1

Or, a = \(\sqrt{L \lambda}\)

B_min  = 2\(2 \sqrt{L \lambda}+2 \sqrt{L \lambda}=4 \sqrt{L \lambda}\)

Hence radius of the spot

b_min = \(\frac{1}{2} \times 4 \sqrt{L \lambda}=2 \sqrt{L \lambda}=\sqrt{4 L \lambda}\)

Polarization multiple choice questions 

Question 53. Unpolarlsed of Intensify I passes through an ideal polariser A, Another identical polariser B is placed behind A. The  Intensity flight beyond B is found to be \(\frac{I}{2}\) Now another Identical polarizer C fa placed between A and B. The Intensity beyond I Is now found to be \(\frac{I}{8}\). The angle between polarisers A and C is

1. 45°
2. 60°
3. 0°
4. 30°

Answer: 1. 45°

It’s The Intensity of unpolarised light after patting through polariser A=\(\frac{I}{2}\). Since the intensity of light does not change after passing through B, the optic axes of polarizers A and B arc parallel,

Let the angle between the optic axes of polarisers A and C be θ. Then the angle between the optic axes polarisers B and C is also θ.

Applying Malus law

⇒ \(\frac{I}{8}=\left(\frac{I}{2} \times \cos ^2 \theta\right) \times \cos ^2 \theta \quad \text { or, } \cos ^2 \theta=\frac{1}{2}\)

Or, = 45°

Question 54. The angular width of the central maximum in a single slit diffraction pattern is 60°. The width of the slit is 1μm. The slit is illuminated by monochromatic plane waves. If another slit of the same width is made near it, Young’s fringes can be observed on a screen placed at a distance of 50 cm from the slits. If the observed fringe width is 1 cm, what is the slit separation distance? (i.e., the distance between the centers of each slit)
Answer: 

The condition for the formation of first minima at points O₁, and O₂

d sin θ = λ

d = width of the slit, θ  = angle of diffraction)

Or, \(\lambda=\frac{d}{2}\left[\text { since } \theta=30^{\circ}\right]\)

Or, \(\lambda=\frac{1 \times 10^{-6}}{2} \mathrm{~m}\)

Or, λ = 5000 A°

The fringe within Young’s double slit experiment

= \(\frac{\lambda D}{d^{\prime}}\)

d’ = \(\frac{5000 \times 10^{-10} \times 0.5}{10^{-2}}\)

= \(2.5 \times 10^{-5}\)m

= 25 μm

Examples of Diffraction Patterns

Question 55. A beam of light of  = 600nm from a distant source falls on a single slit 1mm wide and the resulting diffraction pattern is observed on a screen 2m away. The distance between the first dark fringes on either side of the central bright fringe is

1. 1.2cmλ
2. 1.2mm
3. 2.4 cm
4. 2.4mm

Answer: 1. 1.2cm

The distance of the n -th minima from the central maxima,

x_n = \(\frac{n f \lambda}{a}\) [ a = slit width ]

x₁ = \(\frac{1 \times 200 \times 600 \times 10^{-7}}{0.1}=\frac{12 \times 10^{-3}}{0.1}\)

= 0. 12 cm

Distance of the first minima from the central maxima

= 2x₁ = 2 × 1.2 = 2.4 mm

Question 56. For a parallel beam of monochromatic light of wavelength A, diffraction is produced by a single slit whose width a is of the order of the wavelength of the light. If D is the distance of the screen from the slit, the width of the central maxima will be

1. \(\frac{2 D \lambda}{a}\)
2. \(\frac{D \lambda}{a}\)
3. \(\frac{D a}{\lambda}\)
4. \(\frac{2 D a}{\lambda}\)

Answer: 1. \(\frac{2 D \lambda}{a}\)

The width ofthe central maximum will be \(\frac{2 D \lambda}{a}\)

Question 57. In a double-slit experiment, the two slits are 1 mm apart and the tire screen is placed 1 m away. A monochromatic light of wavelength 500 nm is used. What will be the width of each slit for obtaining ten maxima of double slit within the central maxima of single slit pattern?

1. 0.2 mm
2. 0.1mm
3. 0.5mm
4. 0.02mm

Answer: 1. 0.2 mm

Width of the interference fringe due to the double slit,

y =\(\frac{D}{d} \lambda\)

D = lm,rf= 1 mm and A = 500 nm

Now, the width of the central bright diffraction band due to the single slit,

y ‘= \(\frac{2 D}{a} \lambda\) a = width of the slit

According to the question

⇒ \(\frac{2 D}{a} \lambda=10 \frac{D}{d} \lambda\)

Or, \(a\frac{2}{10} d=\frac{2}{10} \times 1 \mathrm{~mm}\)

= 0.2mm

Polarization multiple choice questions 

Question 58. A parallel beam of light of wavelength A incident normally on a single slit of width d. Diffraction bands are obtained on a screen placed at a distance D from the slit. The second dark band from the central bright band will be at a distance given by

1.  \(\frac{2 \lambda D}{d}\)
2. λdD
3. \(\frac{\lambda D}{2 d}\)
4. \(\frac{2 \lambda d}{D}\)

Answer: 1. \(\frac{2 \lambda D}{d}\)

The distance of n -th dark band from the central bright band = \(\frac{n \lambda D}{d}\)

For n = 2 distance = \(\frac{2 \lambda D}{d}\)

WBCHSE Class 12 Physics MCQs

LightWave And Interference Of Light Multiple Choice Question and Answers

Question 1. In Young’s double-slit experiment, the source is white light If one of the slits is covered by a red filter and the other by a blue filter, then

1. There shall be alternate interference patterns of red and screen blue
2. There shall be an interference pattern for red distinct from that for blue
3. There shall be no interference fringes
4. There shall be an interference pattern for red mixing

Answer: 3

Question 2. A standard two-slit arrangement with slits S₁, S₂. P₁ and P₂ are two minima points on either side of P. At P₂ on the screen, there is a hole and behind P₂ is a second two-slit arrangement with slits S₃, S₄ and a second screen B behind them.

1. There would be no interference pattern on B but it would be lighted ‘
2. The screen B would be totally dark
3. There would be a single bright point on the screen BC
4. Thre would be a regular two-slit pattern on the screen B

Answer: 4

Question 3. Two sources S₁ and S₂ of intensity I₁ and I₂ are placed in front of a screen. The pattern of intensity distribution seen in the central position is given in. Which of the following statements are true

Read and Learn More Class 12 Physics Multiple Choice Questions

1. S₁ and S₂ have the same intensities
2. S₁ and S₂ have a constant phase difference
3. S₁ and S₂ have the same phase
4. S₁ and S₂ have the same wavelength

Answer: 1, 2, and 3

WBBSE Class 12 Light Wave MCQs

Question 4. For light diverging from a point source

1. The wavelength is spherical
2. The intensity decreases in proportion to the distance squared
3. The wavefront is parabolic
4. The intensity of the wavefront does not depend on the distance

Answer: 1,2

1. By Huygens’ wave theory of light, we cannot explain the phenomenon of

1. Interference
2. Diffraction
3. Photoelectric effect
4. Polarisation

Answer: 3. Photoelectric effect

2. By a monochromatic wave, we mean

1. A single ray
2. A single ray of a single colour
3. Wave having a single wavelength
4. Many rays of a single colour

Answer: 3. Wave having a single wavelength

3. If the distance between a point source and the screen is doubled, then the intensity of light on the screen will become

1. Four
2. Double
3. Half
4. One fourth

Answer: 4. One fourth

WBCHSE class 12 physics MCQs

Question 5. Spherical wavefronts, emanating from a point source, strike a plane reflecting the source. What will happen to these wavefronts immediately after reflection?

1. They will remain spherical with the same curvature
2. They will become plane wavefronts
3. They will remain spherical with the same curvature, but a sign of curvature reversed
4. They will remain spherical but with different curvature both in magnitude and sign

Answer: 3. They will remain spherical but with different curvature both in magnitude and sign

Short Answer Questions on Wave Interference

Question 6. The maximum and minimum intensities of the resultant wave due to the superposition of two waves having intensities I and 4I will be

1. 5I and 3I
2. 9I and I
3. 9I and 3I
4. 5I and I

Answer: 2.  9I and I

Question 7. If the ratio of maximum and minimum intensities of the interference fringes obtained in Young’s double slit experiment is 4: 1, then the ratio of the amplitudes of the two coherent sources will be

1. 4: 1
2. 3:1
3. 2:1
4. 1:1

Answer: 2. 3:1

Question 8. If the waves coming out from two sources of light having intensities I and 4I undergo interference, then the intensity at the points in the region of superposition where phase difference becomes \(\pi / 2\) is 

1. I
2. 3I
3. 5I
4. \(\frac{5 I}{2}\)

Answer: 3. 5I

Question 9. Two monochromatic lights coming out from two coherent sources can produce constructive interference when their phase difference becomes

1. \(\frac{3 \pi}{2}\)
2. 2pi
3. pi
4. \(\frac{\pi}{2}\)

Common MCQs on Young’s Double Slit Experiment

Question 10. Two waves of intensities I₁ and I₂, cross a place in direction and same time. The summation of maximum and minimum intensities is

1. I₁+I₂
2. \(\left(\sqrt{I_1}+\sqrt{I_2}\right)^2\)
3. \(\left(\sqrt{I_1}-\sqrt{I_2}\right)^2\)
4. 2(I₁+I₂)

Answer: 4. 2(I₁+I₂)

Question 11. The energy in the interference fringe

1. Produced in bright band
2. Destroyed in the dark band
3. Remainsto bright conserved, band only changes place from a dark band
4. All the above

Answer: 3

Question 12. The ratio of amplitudes of two waves, emitted from two coherent sources is 2: 1. If these two waves get superposed, the ratio of maximum and minimum intensity will be

1. 2
2. 4
3. 9
4. 18

Answer: 3. 9

Optical instruments class 12 MCQs 

Question 13. A double slit experiment where P and Q are the slits. The path lengths PX and QX are nλ and (n +2)λ A respectively. Taking the central fringe as zero, what is formed at XI (n is a whole number and A is wavelength)

1. First bright
2. First dark
3. Second Bright
4. Second dark

Answer: 3. Second Bright

Practice MCQs on Huygens’ Principle

Question 14. If two waves represented by y₁= 4 sin ωt and y₂ = 3 sin\(\left(\omega t+\frac{\pi}{3}\right)\) interfere at a point, the amplitude of the resulting wave will about

1. 7
2. 6
3. 5
4. 3.5

Answer: 2. 6

Question 15. Maximum intensity in Young’s double-slit experiment is 70. If one slit is closed, the intensity would be

1. \(\frac{I_0}{4}\)
2. \(\frac{I_0}{3}\)
3. \(\frac{I_0}{2}\)

Answer: 2. \(\frac{I_0}{4}\)

Question 16. If white light is used instead of monochromatic light in Young’s double slit experiment, then what change in the fringe width will be observed?

1. Interference fringes will disappear
2. No change in interference fringes will occur
3. Interference fringes become colourful
4. The central line of the interference fringe will be of yellow colour

Answer: 3. Interference fringes become colourful

Question 17. If the interference fringe width of the dark band is β₁ and that of the bright band is β₂ then

1. 2β₁ = β₂
2. 2β₂ = β₁
3. β₁ = β₂
4. β₁+3 β₂ = 1

Answer: 3. β₁ = β₂

Optical instruments class 12 MCQs 

Question 18. The distance between the first bright band and the first dark band in interference fringe [symbols have usual meaning ] is

1. \(\frac{\lambda D}{2 d}\)
2. \(\frac{\lambda D}{d}\)
3. \(\frac{\lambda D}{d}\)
4. \(\frac{2 \lambda D}{d}\)

Answer: 3. \(\frac{\lambda D}{d}\)

Question 19. In an ideal double slit experiment, when a glass plate, having a refractive index of 1.5 and thickness t, is placed in the path of interfered light rays of wavelength A then the intensity remains unchanged at the place where the cen¬ tral maximum was previously formed. Then the minimum thickness of the glass plate is

1. 2 λ
2. \(\frac{2 \lambda}{3}\)
3. \(\frac{\lambda}{3}\)
4. λ

Answer: 1. 2 λ

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Question 20. In Young’s double-slit experiment, an electron ray is used. If the velocity of the electron is increased, the fringe width

1. Will Increase
2. Will decrease
3. Will remain the same
4. Fringes will not be visible

Answer: 2. Will decrease

Question 21. In Young’s double slit experiment, the fringe width is found to be 0.4mm. Now if the experiment is conducted by immersing the whole experiment set-up is water (p = 1.33), the magnitude of fringe width would be

1. 0.25mm
2. 0.30mm
3. 0.40mm
4. 0.53 mm

Answer: 4. 0.53 mm

Question 22. In Young’s double slit experiment, the fringe width is found to be 0.3 mm. Now a thin glass plate of refracting index 1.5 is placed in the path of any one of light rays coming from the slits, then the width of the fringe will be

1. 0
2. 0.3 mm
3. 0.45 mm
4. 0.15 mm

Answer: 2. 0.3 mm

Important Definitions in Light Waves and Interference

Question 23. In Young’s double-slit experiment with monochromatic light, interference fringes are obtained on a screen placed at some distance from the slits. If the screen is moved by 5 × 10^-15 m towards the slit, the change in fringe width is 3× 10^-5m. If the distance between the slits is 10^-3 m, the wavelength of the light used is

1. 6000 A°
2. 5000 A°
3. 3000 A°
4. 4500 A°

Answer: 1. 6000 A°

Question 24. In Young’s double slit experiment, the magnitude of the fringe width is β.  If the whole set-up of the experiment is immersed in a liquid of refractive index p., then the magnitude of the fringe width will be

1. μ β
2. \(\frac{\beta}{\mu+1}\)
3. \(\frac{\beta}{\mu-1}\)
4. \(\frac{\beta}{\mu}\)

Answer: \(\frac{\beta}{\mu}\)

Question 25. In Young’s double Slit experiment, the distance between the slits is d. An Interference fringe is obtained on a screen placed at a distance D from the slits. A dark band is noticed just opposite to one of the slits. The wavelength of the light used is

1. \(\frac{D^2}{2 d}\)
2. \(\frac{d^2}{2 D}\)
3. \(\frac{D^2}{d}\)
4. \(\frac{d^2}{D}\)

Answer: 4. \(\frac{d^2}{D}\)

Question 26. Young’s double-slit experiment is conducted using green, red and blue lights in succession. The magnitude of fringe widths obtained are β_G , β_R, β_B, respectively. The observations will be
Answer:

1. β_G > β_B> β_R
2. β_B>β_G > β_R
3.  β_R >β_B>β_G
4.  β_R >β_G >β_B

Answer: 4. β_R >β_G >β_B

Question 27. What is not true for interference oflight?

1. The two sources must be coherent this is a very important condition
2. The intensity of the dark band of the interference fringe may not be zero
3. In interference, energy is not destroyed, only change of place takes place
4. The intensity of the dark band of the interference fringe must be zero

Answer: 4

Question 28. The fringe width of an interference fringe is y for the two given waves.If the frequency of the source becomes double then the fringe width would have been

1. \(\frac{1}{2}\) y
2. y
3. 2y
4. \(\frac{3}{2}\) y

Answer: 1. Answer:

Question 29. For red and blue colours oflight, the distance of mth interfering fringe in Young’s double slit experiment from the central maxima be X_mr and X_mb. Correct relation is

1. X_mr>X_mb
2. X_mr>X_mb
3. X _mr= X_mb
4. X _mr+ X_mb= 0

Answer: 1. X_mr>X_mb

Class 12 physics optical instruments questions 

Question 30. Two sources of wave S₁ and S₂, having zero phase difference, coming from an isolated light source, produce interference. If the common wavelength of the two sources be A then it is observed that a destructive interference has taken place at point P. The value of (S₂P-S₁P) is

1. \(\frac{3 \lambda}{4}\)
2. \(\frac{11 \lambda}{2}\)
3. 2 λ
4. 5 λ

Answer:  2. \(\frac{11 \lambda}{2}\)

Question 31. For the wavelength λ₁ in Young’s double slit experiment, the seventh bright spot is situated at a distance d₁ from the central bright spot. In the same experiment for the same number of bright spot and for the different wavelength λ₂, the distance between two bright spots is d₂. Now the value

1. \(\frac{\lambda_1}{\lambda_2}\)
2. \(\frac{\lambda_2}{\lambda_1}\)
3. \(\frac{\lambda_1^2}{\lambda_2^2}\)
4. \(\frac{\lambda_2^2}{\lambda_1^2}\)

Answer: 1. \(\frac{\lambda_1}{\lambda_2}\)

Question 32. In Young’s double-slit experiment, the distance between the two slits is d and the wavelength of the light used is A. The angular width of the fringe

1. \(\frac{d}{\lambda}\)
2. \(\frac{\lambda}{d}\)
3. \(\frac{2 \lambda}{d}\)
4. \(\frac{\lambda}{2 d}\)

Answer: 4. \(\frac{\lambda}{2 d}\)

Examples of Applications of Light Wave Interference

Question 33. In Young’s double slit experiment, the two slits are d distance apart. The interference pattern is observed on a screen at a distance D from the slits. A dark fringe is observed on the screen directly opposite to one of the slits. The wave¬ length oflight is

1. \(\frac{D^2}{2 d}\)
2. \(\frac{d^2}{2 D}\)
3. \(\frac{D^2}{d}\)
4. \(\frac{d^2}{D}\)

Answer: 4. \(\frac{d^2}{D}\)

Question 34. Which of the following properties of light support the nature of light?

1. Light obeys laws reflection
2. Tight shows interference
3. Tight shows photoelectric effect
4. The speed of light in water is smaller than that in

Answer: 2 And 4

Question 35. Huygens’ principle of secondary wavelets may be used to

1. Find the specific of tight in the vacuum
2. Explain the particle behaviour of light
3. Find the subsequent position of a wavefront
4. Explain Snell’s law

Answer: 2 And 4

Question 36. When light travels from air to glass, a change occurs in its

1. Wavelength
2. Frequency
3. Speed
4. Amplitude

Answer: 1, 3, and 4

Question 37. White light is used in Young’s double-slit experiment. The separation between the slits is b and the screen is at a distance d (d >> b) from the die slits. At a point on the screen directly in front of the sllÿs, certain wavefronts are missing some of these wavelength are.

1. \(\lambda=\frac{b^2}{d}\)
2. \(\lambda=\frac{2 b^2}{d}\)
3. \(\lambda=\frac{b^2}{3 d}\)
4. \(\lambda=\frac{2 b^2}{3 d}\)

Answer: 1, 3

Class 12 physics optical instruments questions 

Question 38. If Young’s double-slit experiment Is performed using white light, then

1. The central fringe will be white
2. No fringe will be completely dark
3. The fringe adjacent to the central one will be red
4. The fringe adjacent to the central node will be violet

Answer: 1,2, 4

Question 39. In the young double slit experiment the light of wavelength λ₂ fringe width is y₁ and the light of wavelength λ₂ fringe width is y₂. If the whole arrangement is dipped into a liquid of refractive index μ, it is found diat for the wavelength λ₁, fringe width becomes y₃. Now the correct relation is

1. \(y_2=y_1 \frac{\lambda_1}{\lambda_2}\)
2. \(y_2=y_1 \frac{\lambda_2}{\lambda_1}\)
3. \(y_3=\frac{y_1}{\mu}\)
4. \(y_3=\mu y_1\)

Answer: 2, 3

Question 40. Interference fringes may be observed due to superposition

1. (1) and (2)
2. (2) and (4)
3. (1) and (3)
4. (3) and (4)

Answer: 1, 4

Four light waves are represented by

(1) y = a₁ sinωt

(2)  y = a₂sin(ωt + e)

(3)  y = a₂sin 2ωt

(4)  y = a₂sin2(ωt + e)

Question 41. In Young’s double slit experiment the ratio of intensities of bright and dark fringes is 9. This means

1. The ratio of the intensities of individual sources is 5: 4
2. The ratio of the intensities of individual sources is 4: 1
3. The ratio of the amplitudes of the light waves is 3: 1
4. The ratio of their amplitudes is

Answer: 2, 4

Question 42. In Young’s double-slit experiment let A and B be the two slits. A thin plate of thickness t and refractive index p is placed in front of A. Let £ be die fringe width. The central maximum will shift

1. Towards A
2. Towards B
3. \(\text { by } t(\mu-1) \frac{\beta}{\lambda}\)
4. \(\text { by } \mu t \frac{\beta}{\lambda}\)

Answer: 1,3

Question 43. In Young’s double-slit experiment the distance of the screen from the plane of the slits is 1.0 m. The wavelength of light used and width of the fringe are 6000 A and 2 mm respectively.

1. Distance between the slits is

1. 0.5 mm
2. 0.4 mm
3. 0.2 mm
4. 0.3 mm

Answer:  4. 0.3 mm

2. If the wavelength of the light used is 4800 A, the width of the fringe will be

1. 6 mm
2. 2.2 mm
3. 2.0 mm
4. 1.8 mm

Answer: 1. 6 mm

Class 12 physics optical instruments questions 

Question 44. In Young’s double slit experiment the slits apart by 3 illuminated with a source of monochromatic light of wavelength 6000 A°. Interference fringes are obtained on a screen at a distance of 1 m from the slits.

1. The width of the fringe will be

1. 0.15 mm
2. 0.2 mm
3. 0.29 mm
4. 0.12 mm

Answer:  2. 0.2 mm

2. If the whole experimental arrangements are immersed in Aater the fringe width will, be

1. 0.15 mm
2. 0.20 m
3. 29 mm
4. 0.12 mm

Answer: 1. 0.15mm

Question 45. When waves from two coherent sources of amplitudes a and b superimpose , the amplitute R of the resultant wave is given by R, \(=\sqrt{a^2+b^2+2 a b \cos \phi}\)

4. where <p is the constant phase angle between the two waves. The resultant intensity l is directly proportional to the square of the amplitude of the resultant wave i.e.,

I∝ (a² + b² + 2abcosΦ)

For constructive interference,

Φ= 2nπ. and I_max =  (a + b)²

For destructive interference,

Φ = – (2n – 1 )π  and I_max =  (a – b)²

If  I₁, I₂ are intensities from two slits of widths ω₁ and ω₂ then

⇒ \(\frac{J_1}{I_2}=\frac{w_1}{w_2}=\frac{a^2}{b^2}\)

Light waves from two coherent sources of intensity ratio 81: 1 produce interference. With the help of the passage choose the most appropriate alternative for each of the following questions.

Real-Life Scenarios in Light Wave Experiments

1. The ratio of amplitudes of two sources is

1. 9:1
2. 81:1
3. 1:9
4. 1: 81

Answer:  1.9:1

The ratio of slit widths of the two sources is

1. 9:1
2. 81: 1
3. 1:9
4. 1: 81

Answer: 2. 81: 1

3. The Die ratio of maxima and minima in the interference I pattern is

1. 9: 1
2. 81:1
3. 1:9
4. 1:81

Answer: 3. 1:9

4. If two slits in Young’s experiment have width ratio 1: 4, the ratio of maximum and minimum intensity in the interference pattern would be

1. 1: 4
2. 1:16
3. 9:1
4. 9:16

Answer: 3. 9:1

Class 12 physics optical instruments questions 

Question  46. Two coherent monochromatic beams of intensities and 41 respectively are superposed. The maximum and minimum intensities in the resulting pattern are

1. 5I and 3I
2. 9I and 3I
3. 4I and I
4. 9I and I

Answer: 4. 9I and I

Question  47. A thin plastic sheet of a refractive index of 1.6 is used to cover one of the slits of a double-slit arrangement. The central point on the screen Is now occupied by what would have been the 7th bright fringe before the plastic was used.If the wavelength of light is 600 nm, what is the thickness (in μm) of the plastic

1. 7
2. 4
3. 8
4. 6

Answer: 1. 7

Fringe width of the seven fringes = 7 × \(\times \frac{D}{2 d} \lambda\)

Again, displacement of the central bright fringe for the plastic sheet of thickness t = \(\frac{D}{2 d}(\mu-1) t\) t

According to the question

⇒ \(7 \frac{D}{2 d} \lambda=\frac{D}{2 d}(\mu-1) t\)

Here, = 1.6 : = 600 nm = 0.6 mA

t = \(\frac{7 \lambda}{\mu-1}=\frac{7 \times 0.6}{1.6-1}\)

= 7m

Question 48. Two monochromatic coherent light beams A and B have intensities L and respectively.If these beams are superposed, the maximum and minimum intensities will be

1. \(\frac{9 L}{4}, \frac{L}{4}\)
2. \(\frac{5 L}{4}, 0\)
3. \(\frac{5 L}{2}, 0\)
4. \(2 L, \frac{L}{2}\)

Answer: 1. \(\frac{9 L}{4}, \frac{L}{4}\)

I_max= \(\left(\sqrt{L}+\sqrt{\frac{L}{4}}\right)^2=\frac{9 L}{4}\)

I_min = \(\left(\sqrt{L}-\sqrt{\frac{L}{4}}\right)^2=\frac{L}{4}\)

Question 49. If Young’s double slit experiment is done with white light, which of the following statements will be true?

1. All the bright fringes will be coloured
2. All the bright fringes will be white
3. The central fringe will be white
4. No stable interference pattern will be visible

Answer: 3.

The central fringe will be white

If Young’s double slit experiment is performed with white light, the centre fringe will be white.

Conceptual Questions on Constructive and Destructive Interference

Question 50. On a hot summer night, the refractive index of air Is smallest near the ground and increases with height from the ground. When a light beam is directed horizontally, the Huygens’s principle leads us to conclude that as it travels, the light beam.

1. Becomeÿ narrower
2. Goes horizontally without any deflection
3. Bends downwards
4. Bends upwards

Answer: 4. Bends upwards

Since air is comparatively lighter near the ground, the wavefronts do not remain parallel any more. Separation between two consecutive wavefronts is greater near the ground than at above. Since a ray of light is always normal to the wavefront, the light beam horizontal to the earth’s surface always bends upwards

Question 51. In Young’s double slit experiment, slits are separated by 0.5 mm, and the screen is placed 150 cm away. A beam of light consisting of two wavelengths, 650 nm and 520 nm is used to obtain interference fringes on the screen. The least distance from the common central maximum to the point where the bright fringes due to both the wavelengths coincide is

1. 1.56 mm
2. 7.8 mm
3. 9.75 mm
4. 15.6 mm

Answer: 2. 7.8 mm

Let the m th bright fringe due to the wavelength 650 nm superposes with the n th bright fringe due to wavelength

Then

⇒ \(\frac{m \lambda_1 D}{d}=\frac{n \lambda_2 D}{d}\)

Or, \(\frac{m}{n}=\frac{\lambda_2}{\lambda_1}=\frac{520}{650}=\frac{4}{5}\)

In case the least distance

m = 4 and n= 5

The least distance = \(\frac{m \lambda D}{d}\)

= \(\frac{4 \times\left(650 \times 10^{-7}\right) \times 150}{0.5 \times 10^{-1}}\) cm

= 7.8 mm

Question 52. In Young’s double-slit experiment, the Intensity of light on the screen whore the path difference μ Is k (λ being the wavelength of light used). The Intensity at a point where the path difference Is \(\frac{\lambda}{4}\) will be

1. k
2. \(\frac{k}{4}\)
3. \(\frac{k}{2}\)
4. Zero

Answer: 3. \(\frac{k}{2}\)

We know, I = \(l_0 \cos ^2 \frac{\pi \delta}{\lambda}\)

When the path difference is λ then, I =  k = I₀ cos²π

When the path difference is \(\frac{\lambda}{4}\) , then I = I \(\cos ^2 \frac{\pi}{4}=\frac{I_0}{2}=\frac{k}{2}\)

Question 53. The intensity at the maximum in Young’s double slit experiment is I₀. The separation between two slits is d = 5λ, where A is the wavelength oflight used in the experiment. What will be the intensity in front of one of the slits on the screen placed at a distance D = 10d?

1. \(\frac{I_0}{4}\)
2. \(\frac{3}{4} I_0\)
3. \(\frac{I_0}{2}\)
4. I₀

Answer: 3. \(\frac{I_0}{2}\)

Path difference = \(\)

Φ = \(\frac{2 \pi}{\lambda} \times \text { path difference }\)

= \(\frac{2 \pi}{\lambda} \times \frac{\lambda}{4}\)

= \(\frac{\pi}{2}\)

Now,I =I \(I_0 \cos ^2 \frac{\phi}{2}\) = \(I_0 \cos ^2 \frac{\pi}{4}\)

= \(\frac{I_0}{2}\)

Ray optics multiple choice questions 

Question 54. In Young’s double slit experiment the separation d between the slits is 2 mm, the wavelength λ of the light used is 5896A° and distance D between the screen and slits Is 100 cm. It is found that the angular width of the fringes is 0.20°. To increase the fringe angular width to 0.21° (with the same λ and D) the separation between the slits needs to be changed to

1. 2.1mm
2. 1.9 mm
3. 1.8mm
4. 1.7mm

Answer: 2.1.9mm

Angular fringe width,

θ = \(\frac{\lambda}{d}\)

= \(\theta_1=\frac{\lambda}{d_1} \quad \text { or, } \theta_1 d_1=\lambda\) ………………………….. (1)

= \(\theta_2=\frac{\lambda}{d_2} \quad \text { or, } \theta_2 d_2=\lambda\) ………………………….. (2)

Dividing (2) by (1) we get,

\(\frac{\theta_2 d_2}{\theta_1 d_1}=\frac{\lambda}{\lambda}\)  = 1

Or, \(d_2=\frac{\theta_1}{\theta_2} d_1=\frac{0.20}{0.21}\) × 2 = 1.9 mm

 

WBCHSE Class 12 Physics MCQs

Optical Instruments Multiple Choice Questions

Question 1. A magnifying glass is used, as the object to be viewed can be brought closer to the eye than the normal near point. This results in

1. A larger angle to be subtended by the object at the eye and hence viewed in greater details
2. The formation of a virtual erect image
3. Increase in the field of view
4. Infinite magnification at the near point

Answer: 1 And 2

Question 2. An astronomical refractive telescope has an objective of a focal length of 20 m and an eyepiece focal length of 2 cm

1. The length of the telescope tube is 20.02 m
2. The magnification is 1000
3. The image for med inverted
4. An objective of a larger aperture.will increase the brightness and reduce the chromatic aberration of the image

Answer: 1, 2 And 3

Read and Learn More Class 12 Physics Multiple Choice Questions

Question 3. A normal eye cannot see the object nearby from a distance of 25 cm, as

1. The focal length of the eye is 25 cm
2. The distance between the eye lens and the retina is 25 cm
3. The eye is unable to adjust the distance between the lens and the retina below a certain limit
4. The eye is unable to adjust the focal length of the eye lens below a certain limit

Answer: 4 The eye is unable to adjust the focal length of the eye lens below a certain limit

Question 4. The ability of the eye to see objects at all distances is called

1. Binocular vision
2. Myopia
3. Hypermetropia
4. Accommodation

Answer: 4 Accommodation

Question 5. The numerical aperture for a human eye is of the order of

1. 1
2. 0.1
3. 0.01
4. 0.001

Answer: 4 0.001

WBBSE Class 12 Optical Instruments MCQs

Question 6. If there had been one eye of the man then

1. The image of the object would have been inverted
2. Visible region would have decreased
3. The image would have not been seen three three-dimensional
4. 2 and 3 both

Answer: 4. 2 and 3 both

WBCHSE class 12 physics MCQs 

Question 7. What type of lens should be used in spectacles for the remedy of myopia?

1. A concave lens whose focal length is equal to the  distance of the far point of the eye defective eye from the lens
2. A convex lens whose focal length is equal to the die distance of the far point of the eye defective eye from the lens
3. A concave lens whose focal length is equal to the distance of the near point of the defective eye from the lens
4. Convex lens whose focal length is equal to the eye distance of the near point of the defective eye from the lens

Answer:  1. A concave lens whose focal length is equal to the  distance of the far point of the eye defective eye from the lens

Question 8. A person who can see things most clearly at a distance of 10cm, requires spectacles to be able to see clearly things at a distance of 30 cm. What should be die focal length of the spectacles?

1. 15 cm (convex)
2. 15 cm (concave)
3. 10 cm
4. Zero

Answer: 1. 15 cm (convex)

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Question 9. The least distance of distinct vision of a man is 45 cm. He uses a lens of focal length 15 cm for reading. The magnification that he gets

1. 4
2. 3
3. 2
4. 1

Answer: 1. 4

Practice MCQs on Telescope and Microscope

Question 10. The image formed by the objective of a compound microscope is

1. Virtual and magnified
2. Virtual and diminished
3. Real and diminished
4. Real and magnified

Answer: 4. Real and magnified

WBCHSE class 12 physics MCQs 

Question 11. To get a large magnification from a compound microscope

1. The focal length of the objective should be large while The focal length of the eyepiece should be small.
2. The focal length of the objective should be small while the focal length of the eyepiece should be large
3. Both of the focal lengths of the objective and the eyepiece should be large
4. Both of the focal lengths of the objective and the eyepiece should be small

Answer:  4. Both of the focal lengths of the objective and the eyepiece should be small

Question 12. The length of the tube of a compound microscope is 21.5 cm. The focal lengths of the objective and the eyepiece are 1.6 cm and 2.1 cm respectively. If the final image is situated at infinity, then the distance of the object from the objective is

1. 3 cm
2. 1.7 cm
3. 6 cm
4. 4.8 cm

Answer: 2. 1.7 cm

Question 13. The angular magnification of a simple microscope can be increased by

1. Increasing the focal length of the lens
2. Increasing the size of the object
3. Increasing the aperture of the lens
4. Increasing the power of the lens

Answer: 4. Increasing the power of the lens

Question 14. The length of the tube of a microscope is 14 cm and its magnifying power for normal eye is 25. The focal length of the eyepiece is 5 cm. The distance of the object from the objective is

1. 2.4 cm
2. 2.5 cm
3. 3.6cm
4. 1.8 cm

Answer: 4. 1.8 cm

Important Definitions in Optical Instruments Q&A

Question 15. Galileo’s telescope has an objective of a focal length of 100 cm and a magnifying power of 50. The distance between the two lenses in normal adjustment is

1. 106 cm
2. 102 cm
3. 92 cm
4. 98 cm

Answer: 4. 98 cm

Question 16. In Galileo’s telescope, the inverted image formed by its. objective serves as a virtual object for its eyepiece. If the eyepiece has to form an inverted and magnified image of the virtual object, the eyepiece has to be a concave lens and it must be so placed that the virtual object falls

1. Within F
2. Between F and 2F
3. At 2F
4. Beyond 2F

Answer: 1. Within F

Question 17. In an astronomical telescope, if the focal lengths of the objective and the eyepiece are f0 and fe respectively, then the imagination of this instrument is almost.
Answer:

1. f_o+f_e
2. f_o × f_e
3. \(\frac{f_o}{f_e}\)
4. ½ (f_o × f_e)

Answer: 3. \(\frac{f_o}{f_e}\)

Optical instruments class 12 MCQs 

Question 18. In the case of the normal focusing of an astronomical telescope, the final image is formed at

1. Focus of the eyepiece
2. Least distance of distinct vision
3. The focus of the objective
4. Infinity

Answer: 4. Focus on the objective

Question 19. The angular magnification of an astronomical telescope will be maximum if the focal lengths of the objective and the eyepiece are respectively

1. 1 m and 5 cm
2. 2 m and 6 cm
3. 3m and 4 cm
4. 4m and 3 cm

Answer: 4. 4m and 3 cm

Question 20. If the focal length of the eyepiece of a telescope is doubled, its magnifying power m will be

1. 2m
2. 3m
3. \(\frac{m}{2}\)
4. 4m

Answer: 3. \(\frac{m}{2}\)

Question 21. When we see an object, the image formed on the retina is

1. Real
2. Virtual
3. Erect
4. Inverted

Answer: 1 And 4

Question 22. In which of the following instruments is the final image erect?

1. Simple microscope
2. Compound microscope
3. Atronomical telescope
4. Galilean telescope

Answer: 1 And 4

Optical instruments class 12 MCQs 

Question 23. Mark the correct options.

1. If the far point increases, the power of the divergent less should be reduced
2. If the near point increases the power of the convergent lens should be reduced
3. If die far point is 1 in away from the eye, divergent less should be used
4. If the near print is 1 m away from the eye, a divergent lens should be used

Answer: 1 And 3

Examples of Applications of Optical Instruments in MCQ Format

Question 24. The focal length of the objective of a compound microscope is f_o and its distance from the eyepiece is L An object is placed at a distance u from the objective. For proper working of the instrument which of the following options are suitable?

1. L<u
2. L>u-
3. f<L<2f
4. L>2f

Answer: 2 and 4

Question 25. A magnifying glass of focal length f used to see an object placed at a distance u from it forms the virtual image at the least distance of distinct vision D. Its magnifying power is given by

1. \(\frac{D}{f}\)
2. 1 – \(\frac{D}{f}\)
3. \(\frac{D}{u}\)
4. 1 + \(\frac{D}{f}\)

Answer: 3 And 4

Question 26. A planet is observed by an astronomical reflecting telescope having an objective of focal length 16 m and eyepiece of focal length 2 cm

1. The distance between the objective and the eyepiece is 16.02 m^-1
2. The angular magnification of the planet is 800
3. The image of the planet is inverted
4. The objective is larger than the eyepiece

Answer: 1,2,3 And 4

Question 27. A compound microscope has an objective lens of focal length 1 cm and an eyepiece of focal length 2.5 cm.

1. If a distinct image of an object situated at a distance of 1.05 cm from the objective is seen, the magnification will be

1. 320
2. 280
3. 220
4. 110

Answer: 3. 220

2. Under the above condition the distance between the two lenses will be

1. 21 cm
2. 32 cm
3. 2.27 cm
4. 23.27 cm

Answer: 4. 23.27 cm

Class 12 physics optical instruments questions 

Question 28. A figure divided into squares, each of size 1.0 mm² is being viewed at a distance of 9.0 cm through a magnifying lens of focal length 10 cm held close to the eye.

1. The magnification produced by the lens will be

1. 10
2. 5
3. 20
4. 25

Answer: 1. 10

2. The area of each square in the virtual image is

1. 1.5 cm²
2. 1 cm²
3. 1.8 cm²
4. 2 cm²

Answer: 2. 1 cm²

3. The angular magnification of the lens is equal to

1. 2.5
2. 3.8
3. 3.5
4. 4.2

Answer: 3. 3.5

Question 29. The focal lengths of the objective and the eyepiece of an astronomical telescope are 140 cm and 5 cm respectively.

1. The magnifying power of the telescope for viewing objects when the final image is formed at the least distance of distinct vision (25 cm) will be

1. 26.5
2. 30
3. 33.6
4. 40

Answer: 1. 26.5

2. When the telescope is in normal adjustment (i.e., the final image is formed at infinity), the magnifying power of the telescope will be

1. 24
2. 28
3. 32
4. 38

Answer: 2. 28

Conceptual MCQs on Light Behavior in Optical Systems

3. When the telescope is in normal adjustment, the separation between the objective and the eyepiece will be

1. 120 cm
2. 140 mm
3. 145 cm
4. 140 cm

Answer: 3. 145 cm

Question 30. In an astronomical telescope, the focal length of the objective is mad

1. Half that of the eyepiece
2. Equal to the eyepiece
3. Shorter than that of the eyepiece
4. Greater than that of the eyepiece

Answer: 4. Greater than that of the eyepiece

Question 31. The intermediate image formed by the objective of a compound microscope is

1. Real, Inverted mid magnified
2. Real erect and magnified
3. Virtual, erect, and magnified
4. Virtual, Inverted, and magnified

Answer: 1. Real, Inverted mid magnified

Class 12 physics optical instruments questions 

Question 32. An observer looks m ti distant tree of height 10 m with a of 20, To the observer the tree telescope of magnifying power appears

1. 10 times taller
2. 10 times nearer
3. 20 times taller
4. 20 times nearer

Answer: 3. 20 times taller

As the tree Is situated at a large distance, the rays coming from It may be taken as parallel. I fence, It can be considered that the tree Is situated at Infinity. In the case of focusing for Infinity, the final Image Is formed at Infinity. Now magnifying power

m = \(\frac{\text { height of the image }}{\text { height of the object }}\)

= 20

Hence, the tree appears 20 times taller

Question 33. If the focal length of the objective lens is Increased then the magnifying power of the

1. The microscope will Increase but that of the telescope decrease
2. Microscope and telescope both will decrease
3. Microscope and telescope both will Increase
4. The microscope will decrease but that of the telescope will Increase

Answer: 4. The microscope will decrease but that of the telescope will Increase

Real-Life Scenarios Involving Optical Instruments

Question 34. An astronomical telescope has an objective and eyepiece of focal lengths of 40 cm and 4 cm respectively. To view 200 cm away from the objective, the lenses separated by a distance

1. 46 cm
2. 50 cm
3. 54cm
4. 37.3 cm

Answer: 3. 54cm

The focal length of the objective, f_o = 40 cm, the focal length of the eyepiece, f_e =4cm, and the distance of the objective

Objective, u_o = -200 an.

Let, the distance of the image from the objective = v_o

∴ \(\frac{1}{v_o}-\frac{1}{u_o}=\frac{1}{f_o}\)

or, \(\frac{1}{v_o}+\frac{1}{200}=\frac{1}{40}\)

or, \(\frac{1}{v_0}=\frac{1}{40}-\frac{1}{200}\)

Or, v_o = 50 cm

If the distance between the eyepiece and the image formed by the tire objective is u_e, the distance of the tire objective from the image

Is v_e and the Image Is formed at Infinity

⇒ \(\frac{1}{v_e}-\frac{1}{u_e}=\frac{1}{f_e}\)

Or, \(\frac{1}{\infty}-\frac{1}{u_e}=\frac{1}{4}\)

u = – 4 cm

Therefore, the distance between the eyepiece and the objective, L =\(\left|v_o\right|+\left|u_{\mathrm{f}}\right|\)

= 50+ 4 = 54 cm

WBCHSE physics optical instruments MCQs 

Question 35. A person has a near point at 60cm. The focal length of spectacles lenses to read at 22cm having glasses separated 2 cm from the eye is

1. 40 cm
2. 10cm
3. 20 cm
4. 30 cm

Answer: 4. 30 cm

Distance of the object from lens, u = -(22-2) = -20cm Image distance, v = -60cm

⇒ \(\frac{1}{f}=\frac{1}{v}-\frac{1}{u}=-\frac{1}{60}+\frac{1}{20}=\frac{1}{30}\)

Hence, f = 30cm

WBCHSE Class 12 Physics MCQs

Refraction Of Light Multiple-Choice Questions

Question 1. A rectangular block of glass ABCD has A p a refractive index 1.6. A pin is placed midway When observed from the face AD, the pin shall

1. Appears to be near A
2. Appears to be near D
3. Appear to be at the centre of AD
4. Not be seen at all

Answer: 4. not be seen at all

Question 2. You are given four sources of light each one providing a light of single colour—red, blue, green and yellow. Suppose the angle of refraction for a beam of yellow light corresponding to a particular angle of incidence at the interface of two media is 90°. Which of the following statements is correct if the source of the yellow light is replaced with other lights without changing the angle of incidence?  

1. The beam of red light would undergo total internal reflection
2. The beam of red light would bend towards the normal while it gets refracted through the second medium
3. The beam of blue light would undergo total internal reflection
4. The beam of green light would bend away from the normal as it gets refracted through the second medium

Answer: 3. The beam of blue light would undergo total internal reflection

Read and Learn More Class 12 Physics Multiple Choice Questions

Question 3. An extended object kept underwater in a deep container appears distorted when seen from near the edge of the container. It so happens because

1. The apparent depth of the points near the edge is less than that of the points away from the edge
2. The angle subtended by the object when submerged in water is less than that when the object is in air
3. Some points away from the edge undergo total internal reflection and become invisible
4. The water in the container acts as a lens and enlarges the object

Answer: 1, 2 And 3

Question 4. If C₁ and C₂ are medium respectively, then the relation of the velocity of light with the refractive index of the medium will be the velocities of light in a vacuum and in a

1. \(\mu=\frac{c_1}{c}\)
2. mu = c1c
3. \(\mu=\frac{c}{c_1}\)
4. μ= C₁  -C₂

Answer: 3 . \(\mu=\frac{c}{c_1}\)

WBBSE Class 12 Refraction of Light MCQs

Question 5. When light moves from glass to air, the property of which remains unchanged is

1. Velocity
2. Wavelength
3. Frequency
4. Amplitude

Answer: 3. Frequency

Question 6. The refractive indices of a medium for two light waves are μ₁ and μ₂ = If μ₁ >μ₂ then which wave will move faster in the medium?

1. First wave
2. Second wave
3. The two waves will move with the same velocity
4. Cannot be said correctly

Answer: Second wave

Question 7. The refractive index of a medium in which the velocity of light is \(2 \times 10^8 \mathrm{~m} \cdot \mathrm{s}^{-1}\) is 

1. 1.4
2. 2.3′
3. 1.5
4. 1.0

Answer: 3. 1.5

Question 8. Wavelength, frequency, velocity and intensity of light when travelling in air is λ, ν, v and I respectively. When the light ray enters the water the values of the above-mentioned properties λ₁, ν₁, v₁ become I₁ and ly respectively. Which relation is correct

1. λ = λ₁
2. ν= ν₁
3. v= v₁
4. I= I₁
   

Answer: 2.  ν= ν₁

Short MCQs on Spherical Lenses

Question 9. In case of refraction of light which of the following phenomena must take place?

1. Change of direction
2. Change of velocity
3. Both  1and 2
4. None of 1,2,3

Answer: 2. Change of velocity

Question 10. If the refractive indices of a particular medium for red and violet light be and respectively, then

1. μ_r >μ_v
2. μ_r <μ_v
3. μ_r = μ_v
4. None of these

Answer: 2. μ_r <μ_v

Question 11. Speed of light through two media of refractive indices and are and respectively

1. u₁= u₁
2. n₁u₁ = n₂ u₂
3. n₁u₂ = n₂ u₁
4. n₁u²₁ = n₂ u²₂

Answer:  n₁u₁ = n₂ u₂

Deviation of a Ray of Light

Question 12. In the case of refraction, angle of deviation is maximum when the angle of incidence is

1. 45°
2. 0°
3. 90°
4. 60°

Answer: 3.90°

Practice MCQs on Image Formation by Lenses

Question 13. In  the case of refraction, the angle of deviation becomes minimum when the angle of incidence becomes

1. 45°
2. 0°
3. 90°
4. 60°

Answer:  2. 0°

Question 14. A diverging beam of rays from a point source S, making a divergent angle, a are incident on a glass slab The angle of incidence of the extreme rays on the two sides is the same. If the thickness of the glass slab is t and the refractive index n, then the E angle of divergence of the emergent rays will be

1. Zero
2. α
3. \(\sin ^{-1} \frac{1}{n}\)
4. 2 \(\sin ^{-1} \frac{1}{n}\)

Answer: 2.α

Question 15. A ray of light is incident on the upper surface of a glass plate of thickness t (fx refractive index of glass). If the angle of incidence i is very small, then the lateral displacement of the emergent ray will be

1. \(\frac{t i \mu}{\mu+1}\)
2. \(\frac{t i(\mu-1)}{\mu}\)
3. \(\frac{t i \mu}{\mu-1}\)
4. \(\frac{t i(\mu+1)}{\mu}\)

Answer: 2 \(\frac{t i(\mu-1)}{\mu}\)

Question 16. A fight ray travels through four adjacent media of refractive μ₁ ,μ₂, μ₃ ,μ₄  indices the bases of the media are parallel if emergent to the incident ray AB Then

1. μ₁ = μ₂
2. μ₂ = μ₃
3. μ₃ = μ₄
4. μ₄ = μ₁

Answer: 4. μ₄ = μ₁

Question 17. Two rays of light are incident normally on die surface of the water \(\frac{4}{3}\). The refractive index of water is|. A glass box is kept j inside the water whose height is h. A ray passes through the box and touches the bottom. The rays 1 and 2 touch the bottom at a time gap of \h’ – height from the top of the glass to the surface of the water; c = speed of light]

1. 0
2. \(\frac{h^{\prime}}{6 c}\)
3. \(\frac{h}{6 c}\)
4. \(\frac{6 h}{c}\)

Answer:  1. 0

Question 18. Let the  XZ plane be the boundary between two transparent media. Medium 1 in Z≥0 has a refractive index of  \(\sqrt{2}\)  and medium with 2  Z<0 has a refractive index of  \(\sqrt{3}\) a ray of light in medium 1 given by the vector A = 6\(\sqrt{3}\)i+8\(\sqrt{2}\) j + 10k  incident on the plane of separation. The angle of refraction in medium 2 is

1. 45°
2. 60°
3. 75°
4. 30°

Answer: 1. 45°

Question 19. The refractive indices of a rarer and a denser medium are μ₁ and μ₂  respectively. The apparent depth of an object, placed in a rarer medium, to an observer situated in a denser medium will be

1. \(\frac{\mu_2}{\mu_1}\)
2. \(\frac{\mu_1}{\mu_2}\)
3. \(\frac{\mu_1}{\mu_2} \times \text { real depth }\)
4. \(\frac{\mu_2}{\mu_1} \times \text { real depth }\)

Answer: 4. \(\frac{\mu_2}{\mu_1} \times \text { real depth }\)

Question 20. There is a point object at the centre of a glass sphere of diameter 12 cm and a refractive index of 1.5. The distance of the virtual image from the surface of the sphere is

1. 4 cm
2. 6 cm
3. 9 cm
4. 12 cm

Answer: 2. 6 cm

WBCHSE class 12 physics MCQs

Question 21. A rectangular glass slab is placed on different alphabets written in different colours. The coloured alphabet which appears to have been raised minimum in comparison with other alphabets is

1. Blue
2. Violet
3. Green
4. Red

Answer: 4. Red

Class 11 Physics	Class 12 Maths	Class 11 Chemistry
NEET Foundation	Class 12 Physics	NEET Physics

Question 22. A container of depth 2d is half-filled with a liquid of refractive index \(\sqrt{2}\) and another half with a liquid of refractive indexμ. The two liquids do not mix with each other. The apparent depth of the inner surface of the bottom of the container will be (neglect the thickness of the bottom of the container)

1. \(\frac{\mu}{d(\mu+2)}\)
2. \(\frac{d(\mu+\sqrt{2})}{\mu \sqrt{2}}\)
3. \(\frac{\sqrt{2} \mu}{d(\mu+\sqrt{2})}\)
4. \(\frac{\mu d}{d+3 \mu}\)

Answer: 2. \(\frac{d(\mu+\sqrt{2})}{\mu \sqrt{2}}\)

Question 23. To a fish underwater, viewing obliquely a fisherman standing on the bank of a lake, the man looks

1. Taller than what he actually is
2. Shorter than he actually is
3. The same height as he actually is
4. Depends on the obliquity

Answer:  1. Taller than what he actually is

Question 24. A transparent cube of 0.21 m edge contains a small air bubble. Its apparent distances when viewed through one face of the cube is 0.10 m and when viewed from the opposite face is 0.04 m. The actual distance of the bubble from the second face of the cube is

1. 0.06 m
2. 0.17 m
3. 0.05 m
4. 0.04 m

Answer: 1. 0.06 m

WBCHSE class 12 physics MCQs

Question 25. in a rectangular glass container five different immiscible transparent liquids A, B, C, D, E are kept according to their respective densities, layerwise. In refractive indices of each liquid has been given. The container is illuminated from one side. A very small piece of glass {fi = 1.61) is gradually dropped into the vessel. During its fall through the layers, the glass piece will become invisible in

1. Liquids A and B
2. The liquid C
3. Liquids D and E
4. All the liquids— A, B, C, D, E

Answer: 2.  The liquid C

Question 26. A lens of a refractive index 1.5 is kept in a medium of refractive index 1.5. The refractive index ofthe lens with respect to the medium will be

1. 1.5
2. 1.5 ×1.5
3. 1
4. 1.5 + 1.5

Answer: 3.1

Question 27. If a ray of light is incident normally on any face of an equi¬ lateral glass prism, then the ray will deviate through an angle

1. 30°
2. 60°
3. 90°
4. 120°

Answer: 1.30°

Important Definitions Related to Refraction MCQs

Question 28. A ray of light is passed from an optically denser medium to a rarer medium. The critical angle for the pair of media is C. The maximum angle of deviation of the ray will be

1. π – C
2. \(\frac{\pi}{2}-C\)
3. 2C
4. \(\frac{\pi}{2}+C\)

Answer: 2. \(\frac{\pi}{2}-C\)

Question 29. Critical angle depends on

1. Colour of light
2. Refractive indices of the two concerned media
3. Colour of light and refractive indices of the two concerned media
4. Wavelength of light

Answer: 2. Colour of light and refractive indices of the two concerned media

Question 30. A ray of light is incident normally on one face of a right-angled isosceles prism. It then grazes the hypotenuse. The refractive index of the material of the prism is

1. 1.33
2. 1. 414
3. 1.5
4. 1.732

Answer: 2. 1. 414

Question 31. Light rays from a point source situated at a depth of h below water can emerge in air through a definite circular section. If the refractive index of water is , the value of the radius of the circular section will be

1. \(\frac{\sqrt{7}}{3} h\)
2. \(\frac{3}{\sqrt{7}} h\)
3. \(\frac{\sqrt{3}}{7} h\)
4. \(\frac{7}{\sqrt{3}} h\)

Answer: 2. \(\frac{3}{\sqrt{7}} h\)

Question 32. A small coin is resting on the bottom of a beaker filled with a liquid. A ray of light from the coin travels up to the surface of the liquid and moves along its surface. What is the velocity of the light in the liquid?

1. \(1.8 \times 10^8 \mathrm{~m} \cdot \mathrm{s}^{-1}\)
2. \(2.4 \times 10^8 \mathrm{~m} \cdot \mathrm{s}^{-1}\)
3. \(3.0 \times 10^8 \mathrm{~m} \cdot \mathrm{s}^{-1}\)
4. \(1.2 \times 10^8 \mathrm{~m} \cdot \mathrm{s}^{-1}\)

Answer:  1. \(1.8 \times 10^8 \mathrm{~m} \cdot \mathrm{s}^{-1}\)

Question 33. How does the surface of water appear to the eye of an observer inside water?.

1. Mirror with a circular hole
2. Mirror with an elliptical hole
3. Mirror without hole
4. Mirror with a square hole

Answer: 1. Mirror with a circular hole

Refraction multiple choice questions 

Question 34. A fish looking up through the water sees the outside world, contained in a circular horizon. If the refractive index of water is \(\frac{4}{3}\)and the fish is 12 cm below the water surface, then the radius of this circle in cm is

1. \(36 \sqrt{7}\)
2. \(\frac{36}{\sqrt{7}}\)
3. \(36 \sqrt{5}\)
4. \(4 \sqrt{5}\)

Answer:  2. \(\frac{36}{\sqrt{7}}\)

Question 35. When a light ray travels from one medium to another and gets refracted, the velocity of light becomes doubled. For total internal reflection in this condition, the value of the critical angle is

1. 30°
2. 90°
3. 60°
4. Cannot be determined

Answer: 1. 30°

Question 36. Light is incident normally on the side AB of a right-angled prism ABC made of glass. A liquid of refractive index j-t is kept above the side AC. The refractive index of the material of 3 the prism is -. What is the value of μ  for which there will be a total internal reflection on AC?

1. \(\mu>\sqrt{3}\)
2. \(\mu<\frac{\sqrt{3}}{2}\)
3. \(\mu<\frac{3 \sqrt{3}}{4}\)
4. \(\mu>\frac{\sqrt{3}}{2}\)

Answer: 3. \(\mu<\frac{3 \sqrt{3}}{4}\)

Question 37. The length of a day due to atmospheric refraction

1. Decreases
2. Increased
3. Remains unchanged
4. Sometimes decreases

Answer: 2.  Increased

Question 38. The twinkling effect of starlight Is due to

1. Total internal reflection
2. High-dense matter of star
3. Constant burning of hydrogen In the star
4. The fluctuating apparent position of the star is slightly different from the actual position of the star

Answer: 4. The fluctuating apparent position of the star is slightly different from the actual position of the star

Question 39. The refractive angle of a prism is 60°. If the prism Is immersed in a liquid then the minimum angle of deviation is 30°. Critical angle of glass with respect to the liquid Is

1. 42°
2. 45°
3. 50°
4. 52°

Answer: 2. 45°

Question 40. A ray of light passes through an isosceles triangle such that the angle of incidence is equal to the angle of emergence. If the angle of emergence and angle of Incidence are both 45° then the angle of deviation will be

1. 15°
2. 75°
3. 90°
4. 150°

Answer: 4. 150°

Refraction multiple choice questions 

Question 41. The refractive index of a prism is \(\sqrt{2}\) and its refractive angle is 60° . If a ray emerges with the minimum angle of deviation then the angle of incidence will be

1. 45°
2. 60°
3. 90°
4. 150°

Answer: 1. 45°

Question 42. A ray of light PQ is incident on the face of an isosceles glass prism kept on a horizontal table. If for the ray PQ the prism is at the position of minimum deviation then

1. α = β
2. α > β
3. α < β
4. α + β = 90

Answer:  1.  α = β

Question 43. The angle of deviation for a thin prism of refractive index 1.5 is 4° for an incident ray. If that prism is dipped in water, then for the same incident ray, the angle of deviation would be \(\text { given } \left.\mu_{\text {water }}=\frac{4}{3}\right]\) 

1. 1°
2. 2°
3. 8°
4. 16°

Answer: 1. 1°

Question 44. A thin prism of angle 15° made of glass of refractive index in glass? μ₁₂= 1.5 is combined with another prism of a glass of refractive index μ₂ = 1-75. The combination of the prisms produces dispersion without deviation, The angle of the second prism should be

1. 5°
2. 7°
3. 10°
4. 12°

Answer: 3. 10°

Examples of Applications of Refraction in Optics

Question 45. A monochromatic beam of light of wavelength λ and frequency f travelling In a vacuum enters a diamond of refractive Index 2.4 t. Then

1. Its wavelength will reduce to \(\frac{\lambda}{2.4}\)
2. Its wavelength will Increase to\(\)
3. Its frequency will reduce to \(\frac{f}{2.4}\)
4. Its wavelength will Increase to \(\frac{c}{2.4}\)

Answer: 1,4

Question 46. A ray of light travelling in a transparent medium falls on a surface separating the medium front air at an angle of Indeuce of 45°. The ray undergoes total (internal reflection. If H Is the refractive Index of the medium with respect to atr, select the possible values) of ft from the following

1. 1.3
2. 1.4
3. 1.5
4. 1.6

Answer:  3,4

Refraction multiple choice questions 

Question 47. For a light ray pawing through a given prism

1. If the angle of incidence is increased, the deviation Increases
2. If the die angle of Incidence is decreased, the deviation increases
3. If the angle of Incidence is either Increased or decreased hum a certain value, the deviation Increases
4. The angle of minimum deviation Is directly piupnttlonal to the angle of the prism if the prism Is thin

Answer:  3,4

Question 48. A bird flies down vertically towards a water surface. To a fish inside the water, vertically below the bird, the bird will appear to

1. Be farther away than its actual distance
2. Be closer than its actual distance
3. Move faster than its actual speed
4. Move slower than its actual speed

Answer:  1,3

Question 49. The refractive index of the material of an equilateral prism is \(\sqrt{2}\)

1. For a ray of light, the minimum angle of deviation is 30°
2. For a ray of light, the minimum angle of deviation is 45°
3. At a 45° angle of incidence, the deviation of a ray becomes the minimum
4. At 60° angle of incidence, the deviation of a ray becomes minimum

Answer:  1,3

Question 50. In a vacuum, the speed of light is ν₀, frequency is n and wavelength is n₀. When a ray travels from one medium to another, the above physical quantities become ν, n and λ respectively, μ is the refractive index of the medium. Which of the following statements is correct?

1. \(n=\frac{n_0}{\mu}\)
2. \(\lambda=\frac{\lambda_0}{\mu}\)
3. \(v=\frac{v_0}{\mu}\)
4. n = n₀

Answer: 2,3, and 4

Question 51. The refractive indices of three media 1,2,  and 3, are (μ₁, μ₂,  and μ₃₎,  respectively  Which of the following statements is correct?

1. Total internal reflection of light ray takes place when it travels from medium 3 to 1
2. The value of the critical angle for refraction of light when it travels from medium 1 to 2 is less than that when it travels from medium 1 to 3
3. The value of critical angle for refraction of light when it travels from medium 1 to 2 is more than that when it travels from medium 1 to 3
4. The possibility of total internal reflection is more when light ray travels from medium 1 to 3 than when it travels from medium 1 to 2

Answer: 1,3 and 4

Question 52. A monochromatic ray of light is incident normally on a refracting face of a prism of angle 30° . The refractive index of the material of the prism is 1.5.

1. The angle of emergence will be

1. 32.5°
2. 20.6°
3. 48.6°
4. 18.6°

Answer: 3.  48.6°

2. The angle of deviation will be

1. 32.5°
2. 20.6°
3. 48.6°
4. 18.6°

Answer: 4. 18.6°

Refraction multiple choice questions 

Question 53. The refractive index ofthe material ofa prism is J? and the refracting angle is 90° .

1. The angle of minimum deviation of the refracted ray by the prism is

1. 30°
2. 35
3. 40°
4. 45°

Answer: 1. 30°

2. The corresponding incidence is

1. 30°
2. 35°
3. 40°
4. 60°

Answer:  4.  60°

3. The limiting angle of incidence for emergent ray is

1. 30°
2. 45°
3. 50°
4. 60°

Answer: 2. 45°

Conceptual MCQs on Focal Length and Magnification

Question 54. A glass slab consists of thin uniform layers of progressively decreasing refractive indices such that the refractive index ofany layer is (i- mA/i. Here fi and Afi denote the refractive index of the 0 th layer and the difference in refractive index sinr sin(90°-/) between any two consecutive layers respectively. The integer m = 0, 1, 2, 3, denotes the numbers of the successive layers. A ray oflight from the 0 th layer enters the 1st layer at an angle of incidence of 30°. After undergoing the m th refraction, the ray emerges parallel to the interface. If μ= 1-5 and Δμ = 0.015 the value of m is

1. 20
2. 30
3. 40
4. 50

Answer:  4. 50

From Snell’s law, we have, for n number of media,

Or,\(\frac{3}{2} \times \frac{1}{2}=(1.5-m \times 0.015) \times 1\)

\(0.015 m=\frac{3}{4}\)

m = \(\frac{3}{4} \times \frac{1000}{15}\)

= 50

Question 55. A ray of light is incident at an angle i on a glass slab of refractive index μ . The angle between reflected and refracted light is 90° . Then the relationship between i and μ is.

1. i = tan-1 (\(\left(\frac{1}{\mu}\right)\)
2. tan i= μ
3. sin i = μ
4. cos i = μ

Answer:  2. tan i= μ

According to

i+ 90+r = 180

Or, r= 90- i

Refractive index = \(\)

= tan i

Question 56. When light is refracted from a surface, which ofits following physical parameters does not change?

1. Velocity
2. Amplitude
3. Frequency
4. Wavelength

Answer:  3. Frequency

When light is refracted from a surface, frequency is the physical parameter that does not change

Question 57. A ray of light strikes a glass plate at an angle of 60°. reflected and refracted rays are perpendicularto each other, the refractive index ofglass is

1. \(\frac{\sqrt{3}}{2}\)
2. \(\frac{3}{2}\)
3. \(\frac{1}{2}\)
4. \(\sqrt{3}\)

Answer: 4. \(\sqrt{3}\)

According to the question,  i= 60°

Hence i+90°+r= 180°

Or, 60°+90°+r= 180°

r = 30°

∴  \(\frac{\sin i}{\sin r}=\mu\)

Or, \(\frac{\sin 60^{\circ}}{\sin 30^{\circ}}=\mu\)

Or, \(\sqrt{3}\)

Question 58. Light travels through a glass plate of thickness t and having refractive index μ. If c is the velocity of light in a vacuum, time taken by the light to travel through this thickness of the glass is

1. \(\frac{t}{\mu c}\)
2. \(\frac{t c}{\mu}\)
3. \(\frac{\mu t}{c}\)
4. μtc

Answer: 3. \(\frac{\mu t}{c}\)

Required time = \(=\frac{\text { thickness of the slab }}{\text { velocity of light in the slab }}\)

= \(\frac{t}{c}=\frac{\mu t}{c}\)

Question 59. Monochromatic light is incident on a glass prism of angle A. If the refractive index ofthe material of the prism is μ , a ray, incident at an angle θ, on the face PQ would get transmitted through the face PR ofthe prism provided.

1. \(\theta>\sin ^{-1}\left[\mu \sin \left(A-\sin ^{-1}\left(\frac{1}{\mu}\right)\right)\right]\)
2. \(\theta<\sin ^{-1}\left[\mu \sin \left(A-\sin ^{-1}\left(\frac{1}{\mu}\right)\right)\right]\)
3. \(\theta>\cos ^{-1}\left[\mu \sin \left(A-\sin ^{-1}\left(\frac{1}{\mu}\right)\right)\right]\)
4. \(\theta<\sin ^{-1}\left[\mu \sin \left(A-\sin ^{-1}\left(\frac{1}{\mu}\right)\right)\right]\)

Answer:  1. \(\theta>\sin ^{-1}\left[\mu \sin \left(A-\sin ^{-1}\left(\frac{1}{\mu}\right)\right)\right]\)

We know \(i_1=\sin ^{-1}\left(\sin A \sqrt{\mu^2-1}-\cos A\right)\)

Here, i₁=θ

Now \(\sqrt{\mu^2-1}-\cos A=\mu\left(\sin A \frac{\sqrt{\mu^2-1}}{\mu}-\cos A \frac{1}{\mu}\right)\)

Where B= \(\sin ^{-1}\left(\frac{1}{\mu}\right)=\cos ^{-1} \frac{\sqrt{\mu^2-1}}{\mu}\)

= \(\mu \sin (A-B)=\mu \sin \left(A-\sin ^{-1} \frac{1}{\mu}\right)\)

The required condition is θ≥ sin^-1 \(\left[\mu \sin \left(A-\sin ^{-1}\left(\frac{1}{\mu}\right)\right)\right]\)

Real-Life Scenarios Involving Refraction Questions

Question 60. In an experiment for determination of the refractive index of glass of prism by i-δ, plot, it was found that a ray incident sin^-1 at angle 35° , suffers a deviation of 40° and that it emerges at angle 79°. In that case which of the following is closest to the maximum possible value ofthe refractive index

1. 1.5
2. 1.6
3. 1.7
4. 1.8

Answer:  1. 1.5

We know, θ = i + e- A

or, 40° = 35° 4- 79° -A or, A = 74 e

Now, the minimum value of the angle of deviation, 8m < 40°

In that case , μ≤ \(\frac{\sin \left(\frac{A+\delta m}{2}\right)}{\sin \frac{A}{2}}\)

Or, μ≤\(\frac{\sin \left(\frac{74^{\circ}+40^{\circ}}{2}\right)}{\sin \frac{74^{\circ}}{2}}\)

Or, 1.39

= 1.5

Question 61. The angle of a prism is A. One of its refracting surfaces Is silvered. Light rays falling at an angle of incidence 2A on the first surface return back through the same path after suffering reflection at the silvered surface. The refractive index μ, of the prism, is

1. 2 sin A
2. 2 cos A
3. \(\frac{1}{2}\) Cos A
4. Tan A

Answer: 2. 2 sin A

According to Snell’s law,

1. sin 2A = sin A

Refractive Index of the material of the prism,

\(=\frac{\sin 2 A}{\sin A}=\frac{2 \cos A \sin A}{\sin A}\)

= 2 cosA

Question 62. The refracting angle of a prism is A, and the refractive index of the material of the prism is cot(A/2). The angle of minimum deviation is

1. 180°- 3A
2. 180°-2A
3. 90°- A-B
4. 180°+ 2A

Answer: 2. 180°-2A

We know \(y=\frac{\sin \frac{A+B_\pi}{2}}{\sin \frac{A}{2}}\)

⇒ \(\sin \frac{A-3 \pi}{2}=p \sin \frac{A}{2}=\cos \frac{A}{2} \sin \frac{A}{2}\)

⇒  cos\(\frac{A}{2}\) = sin(90- \(\frac{A}{2}\)

⇒ \(\frac{A-3}{2} \pi=90^2-\frac{A}{2}\)

Question 63. The angle of incidence for a ray of light at a refracting surface of a prism is 45 the angle of prism is 60 If the ray suffer minimum deviation through the prism the angle of minimum deviation and refractive index of the material of the prism respectively are:

1. 30, \(\sqrt{2}\)
2. 45, \(\sqrt{2}\)
3. 30, \(\frac{1}{\sqrt{2}}\)
4. 45, \(\frac{1}{\sqrt{2}}\)

Answer:  1. 30, \(\sqrt{2}\)

Minimum Angle of deviation

δ_m = 2i – A = 2 × 45° – 60 °= 30°

For minuimu8m deviation r₁ =  \(\frac{A}{2}=\frac{60^{\circ}}{2}\) = 30°

= 30°

For refraction at point p

sin 45 = sin 30°

Or , μ = \(\sqrt{2}\)

Question 64. If the angle of a prism is 60’ and angle of minimumdeviation is 40’, then the angle of refraction will be

1. 4°
2. 30°
3. 20°
4. 3°

Answer:  2. 30°

When angle ofdeviation is minimum, the angle of refraction becomes r = \(r\frac{A}{2}\)

r = \(\frac{60^{\circ}}{2}\) = 30

Question 65. The refractive index ofthe material of a prism is Jl and the angle of the prism is 30’. One of the two refracting surfaces of the prism is made a mirror inwards, by silver coating. A beam of monochromatic light entering the prism from the other face will retrace its path (after reflection from the silvered surface) if Its angle ofincidence on the prism is

1. 30°
2. 45°
3. 60°
4. Zero

Answer: 2. 45°

Applying Snells law at point M,

\(\frac{\sin i}{\sin 30^{\circ}}=\frac{\sqrt{2}}{1}\)

Or, \(\sqrt{2} \times \frac{1}{2}\)

= \(\frac{1}{\sqrt{2}}\)

i = 45°

WBCHSE Class 12 Physics MCQs

Reflection Of Light Multiple-Choice Questions

Question 1. A ray of light is incident on a reflecting surface making an angle of 50° with the surface. Due to reflection, the angle of deviation i

1. 90°
2. 80°
3. 100°
4. 110°

Answer: 100°

Question 2. To form three Images of an object, the angle at which two plane mirrors should be inclined to each other is

1. 30°
2. 60°
3. 90°
4. 120°

Answer: 3. 90°

Question 3. A man moves toward a plane mirror with a velocity 15 m.s^-1 The velocity of the image with respect to the man is

1. 15 m s^-1
2. 30 m-s^-1
3. 35 5 m s^-1
4. 20 m s^-1

Answer: 2. 30 m-s^-1

Read and Learn More Class 12 Physics Multiple Choice Questions

Question 4. If two plane mirrors are inclined at an angle of 45° and an object is placed in between them, the number of images formed will 

1. 7
2. 8
3. 14
4. 4

Answer: 1. 7

WBCHSE class 12 physics MCQs

Question 5. A small object’ is situated at a distance of 10 cm in front of a plane mirror. If you stand behind the object at a distance of 30 cm from the mirror and look at the image, then to the distance of the image will be

1. 60 cm
2. 20 cm40 cm
3. 80 cm

Answer: 3. 40 cm

WBBSE Class 12 Reflection of Light MCQs

Question 6. The reflected rays at a plane mirror can form a real image

1. If the incident rays at the mirror are convergent
2. If the incident rays at the mirror are divergent
3. Under no circumstances
4. If the object is situated very close to the mirror

Answer: 1. If the incident rays at the mirror are convergent

Question 7. Light waves coming from the air are reflected at the surface of the separation of air and glass. The phase change of the reflected wave is equal to

1. Zero
2. π/2
3. π
4. 2π

Answer: 1. Zero

Question 8. In the dial of a clock, there are marks instead of numbers. The reading of time appears 10:10 when the clock is seen through a plane mirror. The real-time of the clock is

1. 9:50
2. 1:50
3. 2:10
4. 10:10

Answer: 2. 1:50

Question 9. A man of height 6 ft wants to see his full image in a plane mirror. The minimum length of the plane mirror required is

1. 6ft
2. 12ft
3. 2ft
4. 3f

Answer: 4. 3f

Question 10. The focal length of a concave mirror is

1. Maximum for red color
2. Maximum for violet color
3. Maximum for the yellow color
4. Equal for all colors

Question 11. Which mirror is to be used to obtain a parallel beam of light from a small lamp

1. Plane mirror
2. Convex mirror
3. Concave mirror
4. Any one of these

Answer: 3. Concave mirror

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Question 12. A concave mirror has a radius of curvature r and focal length f. An object is placed at a distance from the mirror. The condition under which a diminished inverted image will be formed is

1. u= r
2. u >f
3. f<u>r 0
4. 0 <u <f

Answer:  2. u >f

Practice MCQs on Plane Mirrors and Image Formation

Question 13. A concave mirror has a radius of curvature of r and a focal length f. An object is placed at a distance from the mirror. The condition under which an inverted image of the same size of the object will be formed is

1. u= r
2. u>= r
3. f<u>r 0
4. 0 <u >f

Answer: 1. u= r

Question 14. A concave mirror has a radius of curvature r and a focal length f. An object is placed at a distance from the mirror. The condition under which a virtual image will be formed is

1. u = f
2. u>r
3. f<u>r
4. 0 <u<f

Answer: 4. 0 <u<f

Question 15. A concave mirror has a radius of curvature r and a focal length f. An object is placed at a distance x from the mirror. The condition under which a magnified erect image will be formed is

1. x = f
2. u>r
3. f<u>r
4. 0 <u<f

Answer: 4. 0 <u<f

Question 16. A mirror forms a real image of unit magnification. The nature of the mirror is

1. Plane
2. Paraboloidal
3. Convex
4. Concave

Answer: 4. Concave

Question 17. If an object is placed at a distance of 40 cm from a concave mirror of focal length 20 cm, then the image will be

1. Virtual and inverted
2. Real and erect
3. Real, inverted, and diminished
4. Real, inverted, and the same size as the object

Answer: 4. Real, inverted, and the same size as the object

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Question 18. A person standing in front of a mirror finds his image larger than himself. This implies that the mirror is

1. Convex
2. Parabolic
3. Plane
4. Concave

Answer: 4.

Important Definitions in Reflection Q&A

Question 19. A convex mirror has a focal length/. An object is placed at a distance from the mirror. The condition under which a diminished image will be formed is

1. u = 0
2. 0<u< ∞
3. u < 0 and |u| <|f|
4. A diminished image will not be formed, in any way

Answer: 2. 0<u< ∞

Question 20. An- object is placed at a distance of 20 cm from a concave mirror of focal length 10 cm. What is the image distance?

1. 20 cm
2. 6.67 cm
3. 10 cm
4. 30 cm

Answer: 1. 20 cm

Question 21. The radius of curvature of a convex mirror is 40 cm and the size of the object is twice the size of the image. The image distance will be

1. 10 cm
2. 20 cm
3. 40 cm
4. 30 cm

Answer: 1. 10cm

Question 22. A concave mirror of focal length / (in air) is immersed in (H s D. The focal length of the concave mirror in water
water will be

1. f
2. 4/3f
3. 3/4 f
4. 7/3f

Answer: 1. f

Question 23. A concave mirror of focal length 15 cm forms an image twice the size of the object. If the image is virtual, the distance of the object is

1. 45 cm
2. 7.5 cm
3. 30 cm
4. 22.5 cm

Answer: 3. 30 cm

WBCHSE class 12 physics MCQs

Question 24. The image of a point object placed at 10 cm from a concave mirror, is formed at a distance of 20 cm from the mirror. If the object is moved toward the mirror by 0.1 cm then the image

1. Will move another 0.41 cm away from the mirror
2. Will move another 0.81 cm away from the mirror
3. Will move towards the mirror by another 0.41 cm
4. Will move towards the mirror by another 0.81 cm

Answer:  1. Will move another 0.41 cm away from the mirror

Examples of Applications of Reflection in Daily Life

Question 25. A small linear object of length b is situated along the axis of a concave mirror at a distance x. If the focal length of the mirror is f, the length of the image will be

1. \(\left(\frac{f}{u-f}\right)\)b
2. \(\left(\frac{f}{u-f}\right)\)² b
3. \(\left(\frac{f}{u-f}\right)\)b²
4. \(\left(\frac{f}{u-f}\right)\)

Answer:  2. \(\left(\frac{f}{u-f}\right)\)² b

Reflection of light multiple choice questions 

Question 26. In the following figure, if the height of an object is  H₁= +2.5 cm, the height of the image H₂ formed is

1. 5cm
2. +5 cm
3. + 7.5 cm
4. – 7.5 cm

Answer:  1. 5cm

Reflection of light class 12 MCQs 

Question 27. If the object distance and image distance from the focus of a spherical mirror of focal length ƒ be x and y respec- respectively, then

1. xy= f
2. xy= f²
3. x/y= f
4. xy = f³

Answer: 2. xy= f²

Question 28. An object is placed at a distance x from the principal focus of a spherical mirror of focal length f. The magnification of the image will be

1. x/f
2. f/x
3. 1+f/x
4. 1- f/x

Answer: 2. f/x

Question 29. If a graph is drawn taking x and y as the object distance and the image distance respectively from the focus of a spherical mirror, the graph will be a

1. Rectangular hyperbola
2. Parabola
3. Circle
4.  Ellipse

Answer:   1. Rectangular hyperbola

Question 30. A rod is standing in front of a mirror on its axis. The virtual image formed by the mirror is smaller in size than the object. The nature of the mirror is

1. Plane
2. Concave
3. Convex
4. Concave or convex

Answer: 3. Convex

Class 12 physics reflection questions

Question 31. An image that indicates positive magnification is

1.  Erect
2. Inverted
3. Larger than the size of the object
4. Smaller than the size of the object

Answer: 2. Inverted

Question 32. An object is approaching a convex mirror. The ratio of the velocity of the object and that of the object image which is m times magnified is

1. \(-\frac{1}{m^2}\)
2. m²
3. -m
4. \(\frac{1}{m}\)

Answer: 1. \(-\frac{1}{m^2}\)

Question 33. A car is fitted with a convex side-view mirror of a focal length of 20 cm. A second car 2.8 m behind the first car is overtaking the first car at a relative speed 15 m s^-1. The speed of the image of the second car as seen in the mirror of the first one is

1. \(\frac{1}{15} \mathrm{~m} \cdot \mathrm{s}^{-1}\)
2. \(10 \mathrm{~m} \cdot \mathrm{s}^{-1}\)
3. \(15 \mathrm{~m} \cdot \mathrm{s}^{-1}\)
4. \(\frac{1}{10} \mathrm{~m} \cdot \mathrm{s}^{-1}\)

Answer: 1. \(\frac{1}{15} \mathrm{~m} \cdot \mathrm{s}^{-1}\)

Reflection of light multiple choice questions 

Question 34. The image of an extended object, placed perpendicular to the principal axis of a mirror will be erect if 

1. The object and the image are both real
2. The object and the image are both virtual
3. The object is real1but the image in virtual
4. The object is virtual but the image is real

Answer:  3 and 4

Conceptual MCQs on Image Characteristics from Mirrors

Question 35. If the light moving in a straight line bends by a small but fixed angle, it may be a case of

1. Reflection
2. Refraction
3. Diffraction
4. Dispersion

Answer: 1 And 2

Question 36. Two points P and Q lie on either side of an axis XY. It is desired to produce an image of P at Q using a spherical mirror with XY as the optic axis. The mirror must be

1. Converging
2. Diverging
3. Positioned to the left of P
4. Positioned to the right of Q

Answer:  1 And  3

Question 37. If a converging beam of light is incident on a concave mirror, the reflected light

1. May form a real image
2. Must form a real image
3. May form a virtual image
4. Maybe a parallel beam

Answer:  1, 3, And 4

Question 38. A point object P moves towards a convex mirror with a constant speed V along its optic axis. The speed of the image

1. Is always < V
2. Maybe >, = or < V depending on the position of P
3. Increases as P comes closer to the mirror
4. Decreases as P comes closer to the mirror

Answer: 1 And 3

Class 12 physics reflection questions

Question 39. The phenomenon involved in the reflection of radio waves by the ionosphere is similar to

1. The reflection of light by a plane mirrors the total internal
2. Reflection of light in air during a mirage
3. Dispersion of light by water molecules during the formation of a rainbow
4. Scattering of light by particles of air

Answer: 2. Reflection of light in the air during a mirage

Question 40. A car is moving at a constant speed of 60 km h^-1 on a straight road. Looking at the rearview mirror, the driver finds that the car following him is at a distance of 100 m and is approaching at a speed of 5 km h^-1. In order to keep track of the car in the rear, the driver begins to glance alternatively at the rear and side mirror of his car every 2s till the other car overtakes. If the two cars were maintaining their speeds, which of the following statements is correct?

1. The speed of the car in the rear is 65 km hlin
2. The side mirror of the car in the rear would appear to approach with a speed of 5 km h^-1 to the driver of the leading car
3. In the rearview mirror, the speed of the approaching car would appear to decrease as the distance between the cars decreases
4. In the rearview mirror, the speed of the approaching car would appear to decrease as the distance between the cars decreases cars decreases

Answer: 4. In the rearview mirror, the speed of the approaching car would appear to decrease as the distance between the cars decreases cars decreases

Question 41. Rays from the sun subtend an angle θ (in radians) at the pole of a concave mirror of focal length f. If the diameter of the sun is D, the diameter of the image of the sun formed by the mirror is. 

1. Dθ
2. 2 Dθ
3. fθ
4. 2fθ
   

Answer:  3. fθ

If the distance of the sun from the mirror is u,

∴ θ = \(\frac{D}{u}\)

Or, u = \(\frac{D}{u}\)

The rays from the sun form a parallel beam so the image is formed on the focal plane.

So image distance, v= f

Magnification, m= \(\frac{\text { diameter of the image }}{\text { diameter of the sun }(D)}\)

= \(\frac{v}{u}\)

= \(\frac{f}{D / \theta}\)

= \(\frac{f \theta}{D}\)

Hence, the diameter of the image  = D. \(\frac{f \theta}{D}\)

= fθ

Reflection of light multiple choice questions 

Question 42. The ratio of the diameter of the sun to the distance between the Earth and the sun is approximately 0.009.  The approximate diameter of the image of the sun formed by a concave spherical mirror of radius of curvature 0.4 m is.

1. 4.5 × 10^-6m
2. 4.0 ×  10^-6m
3. 3.6×  10^-3m
4. 1.8×  10^-3m

Answer: 4. 1.8×  10^-3m

From the  \(\frac{d}{D}=\frac{f}{x}\)

Or d= \(\frac{f D}{x}\) = 0.009 \(\left(\frac{0.4}{2}\right)\)

= 1.8 × 10^-3 m

Rays from the sun subtend an angle (in radians) at the pole of a concave mirror of focal length f. If the diameter of the sun is D, the diameter of the image of the sun formed by the.

Class 12 physics reflection questions

Question 43.  A point source is placed at coordinates (0, 1) in the XY plane. A ray of light from the source is reflected on a plane mirror placed along the X-axis and perpendicular to the XY plane. The reflected ray passes through the point (3, 3). What is the path length of the ray from (0, 1) to ∞

1. 5
2. \(\sqrt{13}\)
3. 2\(\sqrt{3}\)
4. 1+ 2\(\sqrt{3}\)

Answer:  1. 5

The mirror is placed along the X-axis and perpendicular to the XY-plane i.e., the plane of the mirror is XZ-plane.

∴ Coordinate of image P’ of point P(0, 1) =(0,-1)

∴  Distance traversed by light from P(0, 1) to Q(3, 3)

= PR+ RQ = P’R + RQ = P’Q

= \(\sqrt{(0-3)^2+(-1-3)^2}\)

= \(\sqrt{9+16}\)

= 5 unit

Real-Life Scenarios Involving Reflection Questions

Question 44. An object is placed at a distance of 25 cm from a concave mirror and a real image is formed by the mirror at a distance of 37.5 cm.

1. The focal length of the mirror is:

1. 30 cm
2. 45 cm
3. 15 cm
4.  40 cm

Answer: 2. 45 cm

2. If the object is moved 15 cm toward the mirror the image distance will be:

1. 20 cm
2. 25 cm
3. 30 cm
4. 45 cm

Answer: 2. 25 cm

3. The magnification of the image will be:

1. 2 times
2. 5 times
3. 4 times
4. 3 times

Answer:  4. 3 times

Question 45. The focal length of a concave mirror is f. A point object is placed at a distance of xf on the right side of the focus.

1. The image will be formed on the right side of the focus at a distance of:

1. f/ x
2. f+xf
3. 2f/x
4. f-xf

Answer: 1. f/ x

WBCHSE physics reflection MCQs 

Question 46. A cube of side 2 m is placed in front of a large concave mirror of focal length 1 m in such a way that the face of the cube is at a distance of 3 cm and the face B at a distance of 5 cm from the mirror.

1. The distance between the images of the faces A and B will be

1. 1.5 m
2. 1.75
3. 0.5 m
4. 0.25 m

Answer: 4. 0.25 m

2. The height of the image of the face A will be:

1. 1m
2. 1.5 m
3. 2 m
4. 3m

Answer: 1. 1m

WBCHSE physics reflection MCQs 

3. The height of the image of the face B will be

1. 1m
2.  0.5 m
3. 1.5 m
4. 3 m

Answer: 2. 0.5 m

Question 47. A concave mirror of radius of curvature 20 cm and a convex mirror of radius of curvature 30 cm are held coaxially face to face at a distance of 40 cm apart. An object of height 5 cm is placed perpendicularly on the common axis at a distance of 15 cm from the concave mirror.

1. If the first reflection occurs in the concave mirror, the final image will be formed at a distance of:

1. 6 cm behind the convex mirror
2. 10 cm behind the convex mirror
3. 4 cm in front of the convex mirror
4. 5 cm in front of the convex mirror

Answer: 1. 6 cm behind the convex mirror

2. The height of the final image will be:

1. 4 cm
2. 5 cm
3. 10 cm
4.  6 cm

Answer: 4. 6 cm

Question 48.  An object is placed at a distance of 40 cm from a concave mirror of a focal length 15 cm. If the object is displaced through a distance of 20 cm towards the mirror, the displacement of the image will be

1. 30 cm towards the mirror
2. 3.6 cm away from the mirror
3. 3.0 cm away from the mirror
4. 36 cm towards the mirror

Answer: 2. 3.6 cm away from the mirror

The first case, \(\frac{1}{f}=\frac{1}{v_1}+\frac{1}{u_1}\)

Or, , \(-\frac{1}{15}=\frac{1}{v_1}-\frac{1}{40}\)

[ u₁ = – 40 cm]

Or, v₁ = -24 cm

In 2nd case, \(\frac{1}{f}=\frac{1}{v_2}+\frac{1}{u_2}\)

Or, , \(-\frac{1}{15}=\frac{1}{v_2}-\frac{1}{20}\)

[ u₂ = – 20 cm]

Or, v₂ = -60 cm

WBCHSE Class 12 Physics MCQs

 Magnetic Properties Of Materials Multiple Choice Question And Answers

Question 1. If a current I flows through a loop of area A and the strength of the pole thus generated is qm, the magnetic moment of the loop is

1. IA
2. IA²
3. q_mA
4. q_mA²

Answer: 1. IA

Question 2. The torque acting on a bar magnet of magnetic moment M in a uniform magnetic field B will be

1. MBsinθ
2. MB/sinθ
3. MBcosθ
4. MB/cosθ

Answer: 1. MBsinθ

Question 3. When a magnet is placed in a uniform magnetic field, it experiences

1. A force but no torque
2. A torque but no force
3. A force and also a torque
4. Neither a force nor a torque

Answer: 2. A torque but no force

Read and Learn More Class 12 Physics Multiple Choice Questions

Question 4. In the case of a bar magnet, lines of magnetic induction

1. Start from die North Pole and end at the South Polo
2. Run continuously through the bar and outside
3. Emerge in circular patios from the middle of the bar
4. Are produced only at the north pole like rays of light from a bulb

Answer: 2. Run continuously through the bar and outside

Question 5. The basic element of magnetism is a

1. North pole
2. South pole
3. Dipole
4. Quadrupole

Answer: 3. Dipole

WBBSE Class 12 Magnetic Properties MCQs

Question 6. While entering a paramagnetic material from the air, the spacing between the magnetic lines of force

1. Remains the same
2. Decreases
3. Increases
4. First increases then decreases

Answer: 2. Decreases

WBCHSE class 12 physics MCQs

Question 7. The magnetic susceptibility of a diamagnetic material is

1. Nearly 1000
2. Slightly greater than 1
3. In between 0 and 1
4. Less than 0

Answer: 4. less than 0

Question 8. Curie temperature is the temperature above which

1. Ferromagnetic material becomes paramagnetic
2. Ferromagnetic material becomes diamagnetic
3. Paramagnetic material becomes diamagnetic
4. Paramagnetic material becomes ferromagnetic

Answer: 1. Ferromagnetic material becomes paramagnetic

Question 9. There are four light-weight-rod samples A, B, C, and D separately suspended by threads. A bar magnet is slowly brought near each sample and the following observations are noted.

1. A is feebly repelled
2. B is feebly attracted
3. C is strongly attracted
4. D remains unaffected

Which one of the following is true?

1. B is of a paramagnetic material
2. C is of a diamagnetic material
3. D is of a ferromagnetic material
4. A is of a non-magnetic material

Answer: 1. B is of a paramagnetic material

Common MCQs on Ferromagnetic Materials

Question 10. At a place on the earth’s surface where the horizontal and vertical components of the earth’s magnetic field are equal,

1. The angle of dip is 0°
2. The angle of dip is 90°
3. The angle of the dip is 45°
4. The angle of dip is 30°

Answer: 3. The angle of dip is 45°

Question 11. At any place in the northern hemisphere of Earth, the value of the angle of the dip

1. Is positive everywhere
2. Is negative everywhere
3. Is zero everywhere
4. May be zero, positive, or negative depending on the position of the place

Answer: 1. Is positive everywhere

Question 12. If the Intensity of the geomagnetic field at a place on the magnetic equator of the earth is 20 A.m^-1, die horizontal component of the geomagnetic intensity there is

1. 28 A m^-1
2. >28 A m^-1
3. <28 A m^-1
4. Zero

Answer: 1. 28 A m^-1

WBCHSE class 12 physics MCQs

Question 13. If the intensity of the geomagnetic field at the magnetic poles of the earth is 32 A.m^-1, the horizontal component of. die geomagnetic field intensity there is

1. 32 A.m^-1
2. >32 A m^-1
3. < 32 A.m^-1
4. Zero

Answer: 4. Zero

Question 14. Two short bar magnets of length 1 cm each have magnetic moments i.20 A.m² and 1.00 A m² respectively. They are placed on a horizontal table parallel to each other with their N poles pointing towards the south. They have a common magnetic equator and are separated by a distance of 20.0 cm. The value of the resultant horizontal magnetic induction at the mid-point O of the line joining their centers is close to (a horizontal component of earth’s magnetic induction is 3.6 x 10^-5Wb.m^-2)

1. 3.50 x 10^-4Wb.m^-2
2. 5.80 x 10^-4Wb m^-2
3. 3.6 x 10^-5Wb.m^-2
4. 2.56 x 10^-4Wb.m^-2

Answer: 4. 2.56 x 10^-4Wb.m^-2

Question 15. The magnetic field of Earth can be modeled by that of a point dipole placed at the center of the Earth. The dipole axis makes an angle of 11.3° with the axis of the earth. At Mumbai, the declination is nearly zero. Then

1. The declination varies between 11.3°W to 11.3° E
2. The least declination is 0°
3. The plane defined by the dipole axis and earth axis passes through Greenwich
4. Declination averaged over the earth must be always negative

Answer: 1. The declination varies between 11.3°W to 11.3° E

WBCHSE class 12 physics MCQs

Question 16. A paramagnetic sample shows a. net magnetization of 8 A m^-1 when placed in an external magnetic field of 0.6T at a temperature of 4K. When the same sample is placed in an external magnetic field of 0.2 T at a temperature of 16K, the magnetization would be

1. \(\frac{32}{3} \mathrm{~A} \cdot \mathrm{m}^{-1}\)
2. \(\frac{2}{3} \mathrm{~A} \cdot \mathrm{m}^{-1}\)
3. 6 a.m.^-1
4. 2.4a.m.^-1

Answer: 2. \(\frac{2}{3} \mathrm{~A} \cdot \mathrm{m}^{-1}\)

∴ \(I \propto \frac{B}{T} \quad ..\frac{I_2}{I_1}=\frac{B_2}{B_1} \times \frac{T_1}{T_2}\)

∴ \(I_2=I_1 \times \frac{B_2}{B_1} \times \frac{T_1}{T_2}=8 \times \frac{0.2}{0.6} \times \frac{4}{16}=\frac{2}{3} \mathrm{~A} \cdot \mathrm{m}^{-1}\)

Question 17. A current-carrying circular loop of radius R is placed on the xy plane with the center at the origin. Half of the loop with x>0 is now bent so that it now lies on the yz plane.

1. The magnitude of the magnetic moment now diminishes
2. The magnetic moment does not change.
3. The magnitude of \(\vec{B}\) at (0, 0, z), (z >> R) increases
4. The magnitude of \(\vec{B}\) at (0, 0, z), (z >> R) is unchanged

Answer: 1. The magnitude of the magnetic moment now diminishes

When the circular loop is placed in the xy plane, then

⇒ \(M=I \times \pi R^2 \text {, along }+ \text { ve } z \text {-axis }\)

When half of the loop is bent on yz. plane, the magnetic moment of this portion of the loop,

⇒ \(M_1=\frac{I \cdot \pi R^2}{2}\), along positive x-axis

The magnetic moment of the other portion of the loop (in the xy plane)

⇒ \(M_2=\frac{I \pi R^2}{2}\), along positive z-axis

∴ Resultant moment,

⇒ \(M^{\prime}=\sqrt{M_1^2+M_2^2}=\frac{\sqrt{2}}{2} \cdot I \pi R^2<M\)

Question 18. A circular current loop of magnetic moment M in an arbitrary orientation in an external magnetic field B. The work done to rotate the loop by 30° about an axis perpendicular to its plane is

1. MB
2. \(\frac{\sqrt{3}}{2} M B\)
3. \(\frac{MB}{2}\)
4. Zero

Answer: 4. Zero

The orientation of the loop does not change due to this 30° rotation.

∴ W = 0.

Important Definitions in Magnetic Properties

Question 19. S is the surface of a lump of magnetic material.

1. Lines of \(\vec{B}\) are necessarily continuous across S.
2. Some lines of \(\vec{B}\) must be discontinuous across S.
3. Lines of \(\vec{H}\) are necessarily continuous across S.
4. Lines of \(\vec{H}\) cannot all be continuous across S.

Answer:

1. Lines of \(\vec{B}\) are necessarily continuous across S.

4. Lines of \(\vec{H}\) cannot all be continuous across S.

⇒ \(\vec{B}\) is necessarily continuous across S. Outside the lump of
magnetic material, \(H=\frac{B}{\mu_0}\) and inside the lump,

⇒ \(H=\frac{B}{\mu}=\frac{B}{\mu_0 \mu_r}\)

Thus lines of \(\vec{H}\) cannot be continuous across S.

Magnetic properties of materials class 12 MCQs 

Question 20. If the earth’s magnetic field is supposed to be due to a magnetic dipole placed at the center of the earth, then the angle of dip at a point on the geographic equator

1. Is always zero
2. Can be zero at specific points
3. Can be positive or negative
4. Depends on existing conditions

Answer:

2. Can be zero at specific points

3. Can be positive or negative

4. Depends on existing conditions

Question 21. A long solenoid has 1000 turns per meter and carries a current of 1 A. It has a soft iron core of μ_r = 1000. The core is heated beyond Curie temperature T_C.

1. The \(\vec{H}\) field in the solenoid is almost unchanged, but the \(\vec{B}\) field reduces drastically.
2. The \(\vec{H}\) and \(\vec{B}\) fields in the solenoid are-, nearly unchanged.
3. The magnetization in the core reverses direction.
4. The magnetization in the core diminishes by a factor of 10⁸.

Answer:

1. The \(\vec{H}\) field in the solenoid is almost unchanged, but the \(\vec{B}\) field reduces drastically.

4. The magnetization in the core diminishes by a factor of 10⁸.

H = nI and B = μ₀ μ_r nI

As μ_r changes B will change but H will hot.

When the core is heated beyond Curie’s point, it behaves like a paramagnetic material.

⇒ \(\frac{\chi_{\text {ferro }}}{\chi_{\text {para }}}=\frac{10^3}{10^{-5}}=10^8\)

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Question 22. The gyromagnetic ratio of an electron in an H-atom according to the Bohr model, is

1. Independent of which orbit it is in
2. Negative
3. Positive.
4. Increases with quantum number n

Answer:

1. Independent of which orbit it is in

2. Negative

Gyromagnetic ratio = \(\frac{e}{2 m}\) = constant

Since e is negative, the gyromagnetic ratio is also negative.

Magnetic properties of materials class 12 MCQs 

Question 23. The magnetic field at a distance × along the axis of a short bar magnet is

1. Inversely proportional to x²
2. Inversely proportional to x³
3. Proportional to the dipole moment of the magnet
4. Twice the field at the same distance along the perpendicular bisector of the axis of the magnet

Answer:

2. Inversely proportional to x³

3. Proportional to the dipole moment of the magnet

4. Twice the field at the same distance along the perpendicular bisector of the axis of the magnet

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Question 24. In the classification of magnetic materials based on their behavior.

1. The magnetic permeability of paramagnets is more than 1
2. The magnetic permeability of diamagnets is negative
3. The magnetic permeability of diamagnets does not depend on temperature
4. The magnetic permeability of paramagnets does not depend on temperature

Answer:

1. Magnetic permeability of paramagnets is more than 1

3. Magnetic permeability of diamagnets does not depend on temperature

Question 25. The magnetic moment of a particle with charge q rotating in a circular orbit of radius r and velocity v is p. Then

1. \(p \propto q\)
2. \(p \propto v\)
3. \(p \propto \frac{1}{r}\)
4. p and the angular momentum of the particle is always in the opposite direction

Answer:

1. \(p \propto q\)

2. \(p \propto v\)

Question 26. At a place on the surface of the earth, the angle of dip is θ, the intensity of the earth’s magnetism is I and the horizontal and vertical components of the earth’s magnetism are H and H’ respectively. Then

1. H’ = Htanθ
2. H’ = Icosθ
3. I = Hsecθ
4. The value of I is infinity both at the magnetic north and south pole

Answer:

1. H’ = Htanθ

3. I = Hsecθ

Question 27. Each molecule of iron or nickel behaves as a magnetic dipole. The origin of this magnetism is

1. The electric charge of the molecular electrons
2. The orbital rotation of the electron about the nucleus
3. The spin of the electrons
4. Orbital motion and spin of the electrons

Answer:

1. Electric charge of the molecular electrons

2. Orbital rotation of the electron about the nucleus

Question 28. The magnetic moment of a straight iron wire of length l is p. It is bent in the shape of a semicircle. Then

1. The magnetic moment is p
2. Magnetic moment is \(\frac{2 p}{\pi}\)
3. Magnetic length is l
4. Magnetic length is \(\frac{l}{\pi}\)

Answer:

2. Magnetic moment is \(\frac{2 p}{\pi}\)

4. Magnetic length is \(\frac{l}{\pi}\)

Examples of Applications of Magnetic Materials

Question 29. Relation between the magnetic field vector \(\vec{B}\) and magnetic intensity \(\vec{H}\) at a point in a magnetic field is \(\vec{B}=\mu \vec{H}\) where μ is the magnetic permeability of the medium in which the point is situated. The magnetic permeability of vacuum μ₀ = 47T x 10^-7 H m^-1. Thus the relative magnetic permeability of the medium \(\mu_r=\frac{\mu}{\mu_0}\). To define the magnetic field at a point in a medium another vector needs to be mentioned, which is magnetization \(\vec{M}\). The magnetic moment per unit volume around a point in a magnetic field is known as the magnetization of the point. In most cases \(\vec{M} \propto \vec{H} \text { or, } \vec{M}=k \vec{H}\); this k is called the magnetic susceptibility of the medium. The relation among these three vectors is expressed as

⇒ \(\vec{B}=\mu_0(\vec{H}+\vec{M})\)

1. The magnetic permeability of a medium is 1.26 x 10^-6H m^-1. The magnetic susceptibility of the medium is

1. -0.03
2. -0.003
3. 0.03
4. 0.003

Answer: 4. 0.003

2. Magnetisation at a point in the medium is 0.002 A.m^-1. The intensity at that point is (in A.m^-1)

1. 2.52 x 10^-9
2. 6 x 10^-6
3. 0.667
4. 1.5

Answer: 3. 0.667

3. Magnetic field at the same point (in Wb.m^-2) is

1. 4.2 x 10^-7
2. 8.4 x 10^-7
3. 4.2 x 10^-6
4. 8.4 x 10^-6

Answer: 2. 8.4 x 10^-7

Magnetic properties of materials class 12 MCQs 

Question 30. When current flows in a closed or almost closed path (helical), it is known as a current loop. Any current loop is equivalent to a magnetic dipole whose dipole moment \(\vec{p}=I \vec{A}\), where I am the current and \(\vec{A}\) is the area vector of the loop. If there are N number of turns instead of one, then magnetic moment \(\vec{p}=N I \vec{A}\). When this loop is placed in an external magnetic field S, the torque acting on the loop, \(\vec{G}=\vec{p} \times \vec{B}\). On the other hand, a small bar magnet is also a magnetic dipole whose length vector is \(\vec{r}\) which is in the south-to-north direction. The magnetic dipole moment of the magnet \(\vec{p}=\overrightarrow{q r}\), where q is the pole strength of the magnet.

The force acting on q placed in an external magnetic field of strength \(\vec{B} \text { is } \vec{F}=q \vec{B}\). The magnetic field (B) due to a bar magnet at a point placed at a distance x from the center of the magnet (x>>r) is given by

⇒ \(B=\frac{\mu_0 q r}{4 \pi x^3} \sqrt{3 \cos ^2 \theta+1}\)

where θ = angle between r and the line joining P with the mid-point of the magnet and = permeability of vacuum or air = 4π x 10⁷ H m^-1.

1. The diameter of a circular conducting coil of 10 turns is 10 cm. A current of 10 mA flows through the coil. The magnetic moment of the coil is

1. 7.85 Am²
2. 0.785 A m²
3. 3.14 x 10³A m²
4. 7.85 x 10^-4 A.m²

Answer: 4. 7.85 x 10^-4 A.m²

2. If the magnetic moment of a 10 cm long bar magnet is equal to that of the above coil, then the strength of the magnet is

1. 7.85 A.m
2. 7.85 x 10^-2 A m
3. 7.85 x 10^-3 Am
4. 7.85 x 10^-5 A.m

Answer: 3. 7.85 x 10^-3 Am

Class 12 physics magnetism questions 

3. The coil in question (1) is placed with its axis along an external magnetic field of 10^-4 Wb.m^-2. What is the torque acting on the coil?

1. Zero
2. 7.85 N.m
3. 7.85 x 10^-4 N.m
4. 7.85 x 10^-8 N.m

Answer: 1. Zero

4. What is the magnetic field developed at a point on the perpendicular bisector of the magnet in question (2) the distance of the point from the magnet is 2 m.

1. 9.8 x 10^-21 Wb.m^-2
2. 1.96 x 10^-11 Wb.m^-2
3. 9.8 x 10^-6 Wb.m^-2
4. 1.96 x 10^-5 Wb.m^-2

Answer: 1. 9.8 x 10^-12 Wb.m^-2

5. The magnetic field at a point 2 m away along the axis of the bar magnet is

1. 9.8 x 10^-12 Wb.m^-2
2. 1.96 x 10^-11 Wb.m^-2
3. 9.8 x 10^-6 Wb.m^-2
4. 1.96 x 10^-5 Wb.m^-2

Answer: 2. 1.96 x 10^-11Wb.m^-2

6. The magnetic dipole moment of a particle of charge Q and mass m revolving in a circular path of radius r with a velocity v is

1. \(\frac{Q_{v r}^m}{m}\)
2. \(\frac{Q}{2 m} v r\)
3. Qvr
4. \(\frac{1}{2} Q v r\)

Answer: 4. \(\frac{1}{2} Q v r\)

Question 31. If current I flow in a coil of area A and the number of turns n, the magnetic moment of the coil is

1. nIA
2. n²IA
3. \(\frac{n I}{A}\)
4. \(\frac{n I}{\sqrt{A}}\)

Answer: 1. nIA

Question 32. The ratio of magnetic intensities at points at equal distances on the end-on position and broad-side-on position of a short bar magnet is

1. 2:1
2. 1: 2
3. 3:1
4. 1: 1

Answer: 1. 2: 1

Question 33. The relative magnetic permeability of a diamagnetic substance is

1. Zero
2. Slightly greater than 1
3. Slightly less than 1
4. Slightly less than zero

Answer: 3. Slightly less than 1

Conceptual Questions on Magnetic Field Strength

Question 34. Two similar bar magnets of magnetic moment M each are attached at a right angle with each other at their ends. The magnetic moment of the system will be.

1. M
2. 2M
3. \(\frac{M}{\sqrt{2}}\)
4. √2M

Answer: 4. √2M

If the length of each magnet is L and the pole strength is m,

M = mL

The effective length of the system of the two magnets,

⇒ \(L^{\prime}=\sqrt{L^2+L^2}=\sqrt{2} L\)

∴ Magnetic moment = m. \(\sqrt{2} L=\sqrt{2} M\)

The option 4 is correct.

Class 12 physics magnetism questions 

Question 35. The intensity of magnetization of a bar magnet is 5.0 x 10⁴ A.m^-1. The magnetic length and the area of the cross-section of the magnet are 12 cm and 1 cm² respectively. The magnitude of the magnetic moment of this bar magnet is (in SI unit)

1. 0.6
2. 1.3
3. 1.24
4. 2.4

Answer: 1. 0.6

The volume of the bar magnet,

V = 0.12 x 1 x 10^-4

= 1.2 x 10^-5m³

The intensity of magnetization,

⇒ \(M=\frac{p_m}{V} \quad\left[p_m=\text { magnetic moment }\right]\)

∴ pm = MV

= 5.0 x 10⁴ x 1.2 x 10^-5

= 0.6 A.m²

Option 1 is correct.

Question 36. An electron in a circular orbit of radius 0.05 nm performs 10¹⁶ revolutions per second. The magnetic moment due to this rotation of electron is (in A.m²)

1. 2.16 x 10^-23
2. 3.21 x 10^-22
3. 3.21 x 10^-24
4. 1.26 x 10^-23

Answer: 4. 1.26 x 10^-23

Magnetic moment

= IA

= qnA

= (1.6 x 10^-19) x 10¹⁶ x π(0.05 x 10^-5)²

= 1.26 x 10^-53 A.m²

The option 4 is correct.

Question 37. If \(\chi\) stands for the magnetic susceptibility of a substance, p for its magnetic permeability, and pQ for the permeability of free space, then

1. For a paramagnetic substance: \(\chi\) > 0, μ >0
2. For a paramagnetic substance: \(\chi\) > 0 , μ > μ₀
3. For a diamagnetic substance: \(\chi\) > 0 , μ < 0
4. For a ferromagnetic substance: \(\chi\) > 1 > μ > μ₀

Answer: 2 and 4 are correct.

For both paramagnetic and ferromagnetic substances, μ > μ₀.

But for paramagnetic substance \(\chi\) > 0 and ferromagnetic substance \(\chi\) > 1.

If 0 < μ < μ₀, the substance is not paramagnetic.

Options 2 and 4 are correct.

Question 38. The coercivity of a small magnet where the ferromagnet gets demagnetized is. 3 x 10³ A.m^-1. The current required to be passed in a solenoid of length 10 cm and number of turns 100 so that the magnet gets demagnetized when inside the solenoid is

1. 6A
2. 30 mA
3. 60 mA
4. 3 A

Answer: 4. 3 A

⇒ \(B=\mu_0 n I \quad \text { or, } \frac{B}{\mu_0}=n I \quad \text { or, } H=\frac{N I}{L} \quad\left[∵ B=\mu_0 H\right]\)

∴ \(I=\frac{H L}{N}=\frac{3 \times 10^3 \times 10 \times 10^{-2}}{100}\)

= 3A

The option 4 is correct.

Question 39. Hysteresis loops for two magnetic materials A and B are given below. These materials are used to make magnets for electric generators, transformer cores, and electromagnet cores.

1. A for electric generators and transformers
2. A for electromagnets and B for electric generators
3. A for transformers and B for electric generators
4. B for electromagnets and transformers

Answer: 4. B for electromagnets and transformers

Class 12 physics magnetism questions 

Question 40. A magnetic needle of magnetic moment 6.7 x 10^-2 A.m² and moment of inertia 7.5 x 10⁶ kg.m^-2 is performing simple harmonic oscillations in a magnetic field of 0.01 T. Time taken for 10 complete oscillations is

1. 6.65 s
2. 8.89 s
3. 6.98 s
4. 8.76 s

Answer: 1. 6.65 s

Time period, \(T=2 \pi \sqrt{\frac{I}{M B}}=2 \pi \sqrt{\frac{7.5 \times 10-6}{6.7 \times 10^{-2} \times 0.01}}\)

= 6.65 x 10^-1

∴ Time taken for 10 complete oscillations

= 6.65 x 10^-1 x 10

= 6.65 s

The option 1 is correct.

Real-Life Scenarios in Magnetism Experiments

Question 41. The dipole moment of a circular loop carrying a current J, is m and the magnetic field at the center of the loop is B₁. When the dipole moment is doubled by keeping the current constant, the magnetic field at the center of the loop is B₂. The ratio \(\frac{B_1}{B_2}\) is

1. √2
2. \(\frac{1}{\sqrt{2}}\)
3. 2
4. √3

Answer: 1. √2

We know, dipole moment, m = IA [I is loop current, A is area of the loop]

∴ m = I x πR² [R is the radius of the loop]

According to the question, in the second case dipole moment,

⇒ \(m^{\prime}=2 m=I \times \pi(\sqrt{2} R)^2\)

∴ In the second case, the radius of the loop will be √2R

⇒ \(\text { Now, } \frac{B_1}{B_2}=\frac{\frac{\mu_0 I}{2 R}}{\frac{\mu_0 I}{2 \times \sqrt{2} R}}=\sqrt{2}\)

Option 1 is correct.

Question 42. The magnetic susceptibility is negative for

1. Paramagnetic material only
2. Ferromagnetic material only
3. Paramagnetic and ferromagnetic materials
4. Diamagnetic material only

Answer: 4. Diamagnetic material only

Question 43. The following figures show an arrangement of bar magnets in different configurations. Each magnet has magnetic dipole moment \(\vec{m}\). Which configuration has the highest net magnetic dipole moment?

Answer: 3.

For the configuration 1, total magnetic dipole moment,

⇒ \(\sqrt{m^2+m^2}\)

= \(\sqrt{2} m\)

= 1.41m

For the configuration 2, total magnetic dipole moment,

= m-m

= 0

For the configuration 3, total magnetic dipole moment,

⇒ \(=\sqrt{m^2+m^2+2 m^2 \cos 30^{\circ}}\)

= 1.93m

For the configuration 4, total magnetic dipole moment,

⇒ \(\sqrt{m^2+m^2+2 m^2 \cos 60^{\circ}}\)

= 1.73m

The option 3 is correct.

Question 44. Two reasons for using soft iron as the material for electromagnets.

1. Low permeability and high retentivity
2. High permeability and low retentivity
3. Low permeability and low retentivity field
4. High permeability and high retentivity

Answer: 2. High permeability and low retentivity

The magnetic permeability of soft iron is very high. So, when current flows through it, the electromagnet becomes very strong.

On the other hand, as its retentivity is low, the magnetism disappears as soon as the current flow is stopped.

The option 2 is correct.

WBCHSE physics magnetic properties MCQs 

Question 45. A thin diamagnetic rod is placed vertically between the poles of an electromagnet. When the current in the electromagnet is switched on, then the diamagnetic rod is pushed up, out of the horizontal magnetic field. Hence the rod gains gravitational potential energy. The work required to do this comes from

1. The lattice structure of the material of the rod
2. The magnetic field
3. The current source
4. The induced electric field due to the changing magnetic field

Answer: 3. The current source

The electrical energy of the electric source is converted into stored gravitational potential energy in the rod.

The option 3 is correct.

WBCHSE Class 12 Physics Electromagnetism MCQs

Electromagnetism Multiple Choice Question And Answers

Question 1. Which of the properties of an isolated north pole, placed at a pointin’ field, is characterized by the direction of the tangent on the magnetic line of force passing through that point?

1. Position
2. Displacement
3. Velocity
4. Acceleration

Answer: 4. Acceleration

Question 2. Magnetic flux is defined as

1. The number of magnetic lines of force passing through a surface
2. The number of magnetic lines of force passing normally through a surface
3. The number of magnetic lines of force passing normally through the unit area of a surface
4. The number of magnetic Hnef-ice pa unit area of a surface

Answer: 2. The number of magnetic lines of force passing normally through a surface

Question 3. Current I is flowing through a vertical long wire in the upward direction. The magnetic field at a point, on the east of the wire is

1. Upwards
2. Towards north
3. Towards south
4. Towards west

Answer: 2. Towards north

Read and Learn More Class 12 Physics Multiple Choice Questions

Question 4. A current of 1 A is flowing through a circular coil of radius 10 cm having N turns. If the magnetic field produced at the center of the coil is 4π x 10^-6 T, what is the value of N?

1. 20
2. 10
3. 2
4. 1

Answer: 3. 2

WBCHSE class 12 physics MCQs

Question 5. Which one le of the following relations expresses Biot-Savart law?

1. \(d \vec{B}=\frac{\mu_0 I}{4 \pi}: \frac{d l \times \vec{r}}{r^2}\)
2. \(d \vec{B}=\frac{\mu_0 I}{4 \pi} \cdot \frac{d \vec{l} \times \vec{r}}{r^3}\)
3. \(d \vec{B}=\frac{\mu_0 I}{4 \pi} \cdot \frac{d \vec{l} \times \vec{r}}{r}\)
4. \(d \vec{B}=\frac{\mu_0 I}{4 \pi} \cdot \frac{d \vec{l} \times \hat{r}}{r}\)

Answer: 2. \(d \vec{B}=\frac{\mu_0 I}{4 \pi} \cdot \frac{d \vec{l} \times \vec{r}}{r^3}\)

WBBSE Class 12 Electromagnetism MCQs

Question 6. If we double the radius of a current-carrying coil keeping the current unchanged, the magnetic field at its centre will

1. Remain Unchanged
2. Become Double
3. Be Halved
4. Become four times

Answer: 3. Be Halved

Question 7. A circular coil of radius r carries a current. It produces magnetic fields B1 at the center of the coil, and B₁ and B₂ at an axial point at a distance r from the center. The ratio of B₁ and B₂ is

1. √2: 1
2. 2: 1
3. 2√2: 1
4. 4: 1

Answer: 3. 2√2: 1

Question 8. A conductor is carrying a current I. The magnitude of the magnetic field at the origin is

1. \(-\frac{\mu_0 I}{4 r}\left(\frac{1}{\pi} \hat{i}+\frac{1}{2} \hat{k}\right)\)
2. \(\frac{\mu_0 I}{4 \pi}\left(\frac{1}{\pi} \hat{i}-\frac{1}{2} \hat{j}\right)\)
3. \(\frac{\mu_0 I}{4 r}\left(\frac{1}{2} \hat{i}-\frac{I}{R} \hat{j}\right) \frac{\mu_0}{4 r}\)
4. \(\frac{\mu_0 I}{4 r}\left(\frac{2}{\pi} \hat{i}+\hat{j}\right)\)

Answer: 1. \(-\frac{\mu_0 I}{4 r}\left(\frac{1}{\pi} \hat{i}+\frac{1}{2} \hat{k}\right)\)

Question 9. The ratio of magnetic fields at the center of a current-carrying coil of radius r and at a distance of 3 r on its axis is

1. V10
2. 2V10
3. 10V10
4. 20V10

Answer: 3. 10V10

WBCHSE class 12 physics MCQs

Question 10. Two wires PQ and QR carry equal currents I. One end of each wire extends to infinity and ∠PQR = θ. The magnitude of the magnetic field at O on the bisector of angle ∠PQR at a distance r from point Q is

1. \(\frac{\mu_0}{4 \pi} \cdot \frac{I}{r} \sin \frac{\theta}{2}\)
2. \(\frac{\mu_0}{4 \pi} \cdot \frac{I}{r} \cot \left(\frac{\theta}{2}\right)\)
3. \(\frac{\mu_0}{4 \pi} \cdot \frac{I}{r} \tan \frac{\theta}{2}\)
4. \(\frac{\mu_0}{2 \pi} \cdot \frac{I}{r}\left(\frac{1+\cos \frac{\theta}{2}}{\sin \frac{\theta}{2}}\right)\)

Answer: 4. \(\frac{\mu_0}{2 \pi} \cdot \frac{I}{r}\left(\frac{1+\cos \frac{\theta}{2}}{\sin \frac{\theta}{2}}\right)\)

Question 11. A vertical straight conductor carries a current vertically upwards. A point P lies to the east of it at a small distance and point Q lies to the west at the same distance. The strength of the magnetic field at P is

1. Greater than that at Q
2. Same as that at Q
3. Less than that at Q
4. Greater or less depends on the strength of the current

Answer: 2. Same as that at Q

Class 12 Physics Chapter 1 Electromagnetism Multiple Choice Questions WBCHSE 

Question 12. The magnitude of the magnetic field at the point P in the arrangement.

1. \(\frac{\mu_0 i}{\sqrt{2} \pi d}\left(1-\frac{1}{\sqrt{2}}\right)\)
2. \(\frac{2 \mu_0 i}{\sqrt{2} \pi d}\)
3. \(\frac{\mu_0 i}{\sqrt{2} \pi d}\)
4. \(\frac{\mu_0 i}{\sqrt{2} \pi d}\left(1+\frac{1}{\sqrt{2}}\right)\)

Answer: 1. \(\frac{\mu_0 i}{\sqrt{2} \pi d}\left(1-\frac{1}{\sqrt{2}}\right)\)

Question 13. The magnetic field at point O due to the current I flowing through the wire, is

1. \(\frac{2 \mu_0 I}{r}\)
2. \(\frac{\mu_0 I}{r}\)
3. \(\frac{\mu_0 I}{2 r}\)
4. \(\frac{\mu_0 I}{4 r}\)

Answer: 4. \(\frac{\mu_0 I}{4 r}\)

Question 14. The magnetic field at the point of intersection of diagonals of a square wire loop of side L carrying a current I is

1. \(\frac{\mu_0 I}{\pi L}\)
2. \(\frac{2 \mu_0 I}{\pi L}\)
3. \(\frac{\sqrt{2} \mu_0 I}{\pi L}\)
4. \(\frac{2 \sqrt{2} \mu_0 I}{\pi}\)

Answer: 4. \(\frac{2 \sqrt{2} \mu_0 I}{\pi}\)

Question 15. A circular coil of radius R carries an electric current i. The magnetic field at a point on the axis at a distance x from, the center of the coil (x >> k) varies as

1. \(\frac{1}{x}\)
2. \(\frac{1}{x^{3 / 2}}\)
3. \(\frac{1}{x^2}\)
4. \(\frac{1}{x^3}\)

Answer: 4. \(\frac{1}{x^3}\)

Class 12 Physics Chapter 1 Electromagnetism Multiple Choice Questions WBCHSE 

Question 16. The magnetic field B within a solenoid of length I with N turns and carrying a current i is given by

1. \(\frac{\mu_0 N i}{e L}\)
2. \(\mu_0 N i\)
3. \(\frac{\mu_0^* N i}{L}\)
4. \(\frac{4 \pi \mu_0 N i}{L}\)

Answer: 3. \(\frac{4 \pi \mu_0 N i}{L}\)

Question 17. In a current-carrying, long solenoid, the field produced inside the solenoid does not depend upon

1. The radius of the solenoid
2. Number of turns per unit length
3. Current flowing through it
4. The medium in which the solenoid is placed

Answer: 1. Radius of the solenoid

Common MCQs on Electromagnetic Theory

Question 18. A charged particle enters a magnetic field \(\vec{B}\) perpendicularly with velocity v and keeps rotating along a circular path of radius r. What will happen if the magnitude of \(\vec{B}\) is increased?

1. v will increase
2. v will decrease
3. r will increase
4. r will decrease

Answer: 4. r will decrease

Question 19. The radius of the circular path described by a charged particle in a magnetic field is directly proportional to the

1. Momentum of the particle
2. The kinetic energy of the particle
3. Amount of charge of the particle
4. Strength of the magnetic field

Answer: 1. Momentum of the particle

Question 20. The magnitude of an electric field along the x-axis is 1 V .m^-1 and in the same region the magnitude of a magnetic field along the y-axis is 10^-6 T . What should be the velocity of an electron in that region so that it will continue to move with uniform velocity along z-axis without suffering any deviation?

1. 10⁶ m .s^-1
2. 10^-6 m.s^-1
3. 2 x 10⁶ m.s^-1
4. 2 x 10^-6 m.s^-1

Answer: 1. 10⁶ m.s^-1

Class 12 Physics Chapter 1 Electromagnetism Multiple Choice Questions WBCHSE

Question 21. A beam of protons projected along a positive X-axis experiences a force, due to a magnetic field, along the negative Y-axis. Then the magnetic field must be

1. Along the z-axis
2. Along the negative z-axis
3. On the xy-plane
4. On the xz-plane

Answer: 4. On the xz -plane

Question 22. A moving electron and a moving proton enter a uniform magnetic field in a direction perpendicular to that of the field. If the radii of their circular orbits are equal, they have the same

1. Velocity
2. Momentum
3. Kinetic energy
4. Charge to mass ratio

Answer: 2. Momentum

Question 23. A particle and a proton having the same momentum enter a region of a uniform magnetic field and move in circular paths. The ratio of the radii of curvature of their circular paths ra/rp in the field is

1. 1
2. 1/4
3. 1/2
4. 4

Answer: 3. 1/2

Question 24. A uniform electric field and a uniform magnetic field are acting along die same direction in a certain region. If an electron is projected in the region with a velocity along the direction of fields, then

1. The electron will turn toward the right
2. The speed of the electron will decrease
3. The speed of the electron will increase
4. The electron will turn toward the left

Answer: 2. The speed of the electron will decrease

Question 25. A particle of charge q moves with a velocity \(\vec{v}=a \hat{i}+b \hat{j}\) in a magnetic field \(\vec{B}=c \hat{i}+d \hat{j}\). The force acting on the particle has magnitude F. Then

1. F = 0 if ad = bc
2. F = 0 if ad = -bc
3. F = 0 if ac = -bd
4. \(F \propto\left(a^2+b^2\right)^{1 / 2} \times\left(c^2+d^2\right)^{1 / 2}\)

Answer: 1. F = 0 if ad = bc

Question 26. A proton, a deuteron, and an a -particle are accelerated by the same potential and then enter a uniform magnetic field perpendicularly. The ratio of radii of their circular paths will be

1. 1:√2:√2
2. 2: 2: 1
3. 1:2:1
4. 1: 1: 1

Answer: 1. 1:√2:√2

Class 12 Physics Chapter 1 Electromagnetism Multiple Choice Questions WBCHSE 

Question 27. Through a straight conducting wire, the current is flowing along a positive z-direction. What should be the direction of the applied magnetic field so that the wire will experience the maximum force?

1. Along the positive or negative z-axis
2. Along any direction, the xz-plane
3. Along any direction on xy-plane
4. Along any direction on Yz-plane

Answer: 3. Along any direction on xy-plane

Electric energy and power class 12 MCQs

Question 28. A conducting circular loop of radius r carries a constant current I. It is placed in a uniform magnetic field \(\vec{B}\) such that \(\vec{B}\) is perpendicular to the plane of the loop. The magnetic force acting on the loop is

1. BIr
2. 2πrIB
3. Zero
4. πrIB

Answer: 3. Zero

Practice Questions on Magnetic Fields and Forces

Question 29. A magnetic field is applied along the positive z-axis. How should a plane conducting loop be placed in this field so that the loop will not experience any torque?

1. On xy-plane
2. On xz-piane
3. On yz -plane
4. Along z-axis

Answer: 1. On xy-plane

Question 30. The path of a charged particle whose motion is perpendicular to a uniform magnetic field is

1. A straight line
2. An ellipse
3. A circle
4. A helix

Answer: 3. A circle

Question 31. Two concentric- coils each of radius 2π. cm are placed at right angles to each other. 3 A and 4A are die currents flowing in the coils. The magnetic induction in weber/m² at the centre of the coils will be (\(\mu_0=4 \pi \times 10^{-7} \mathrm{H} \cdot \mathrm{m}^{-1}\))

1. 5 x 10^-5
2. 7 x 10^-5
3. 12 x 10^-5
4. 10^-5

Answer: 1. 5 x 10^-5

Question 32. A square loop carrying a steady current I is placed on a horizontal plane near a long straight conductor carrying a steady current I1 at a distance. The loop will experience

1. A net attractive force toward the conductor
2. A net repulsive force away from the conductor
3. A net torque acting upwards perpendicular to the horizontal plane
4. A net torque acting downwards normal to the

Answer: 1. A net attractive force toward the conductor

WBCHSE Physics Electromagnetism Chapter MCQs 

Electric energy and power class 12 MCQs

Question 33. A square loop of side a is placed at a distance from a long wire carrying a current. If the loop carries a current I₂, then the nature of force and its magnitude is

1. \(\text { attractive, } \frac{\mu_0 I_1 I_2}{2 \pi a}\)
2. \(\text { attractive, } \frac{\mu_0 I_1 I_2}{4 \pi}\)
3. \(\text { repulsive, } \frac{\mu_0 I_1 I_2}{4 \pi}\)
4. \(\text { repulsive, } \frac{\mu_0 I_1 I_2}{4 \pi a}\)

Answer: 2. \(\text { attractive, } \frac{\mu_0 I_1 I_2}{4 \pi}\)

Question 34. For 1 A current, a galvanometer shows its full-scale deflection. If a resistance of 800Ω is connected in series, it is converted into a voltmeter of range 0-1000 V. What is the resistance of the galvanometer?

1. 50Ω
2. 100Ω
3. 200Ω
4. 800Ω

Answer: 3. 200Ω

Question 35. In an ammeter, 0.5% of the main current passes through the galvanometer. If the resistance of the galvanometer is G, the resistance of the ammeter will be

1. \(\frac{G}{200}\)
2. \(\frac{G}{199}\)
3. 200 G
4. 199 G

Answer: 1. \(\frac{G}{200}\)

Question 36. Two charged particles traverse identical helical paths completely opposite sense in a uniform magnetic field, \(\vec{B}=B_0 \hat{k}\)

1. They have equal z-components of momenta
2. They must have equal charge
3. They necessarily represent particle-antiparticle pair
4. The charge to mass ratio satisfy \(\left(\frac{q}{m}\right)_1+\left(\frac{q}{m}\right)_2=0\)

Answer: 4. The charge to mass ratio satisfy \(\left(\frac{q}{m}\right)_1+\left(\frac{q}{m}\right)_2=0\)

The radius of the helical path, \(r=\frac{\nu \sin \theta}{\left(\frac{q}{m}\right) B}\)

and the displacement of the charged particle along the magnetic field is one rotation

⇒ \(\frac{2 \pi \nu \cos \theta}{\left(\frac{q}{m}\right) B}\)

Two charged particles will traverse identical helical paths in the opposite sense is \(\frac{q}{m}\) for both the particles is same and opposite.

WBCHSE Physics Electromagnetism Chapter MCQs 

Question 37. An electron is projected with uniform velocity along the axis of a current carrying a long solenoid. Which one of the following is true?

1. The electron will be accelerated along the axis
2. The electron path will be circular about the axis
3. The electron will experience a force at 45° to the axis and execute a helical path
4. The electron will continue to move with uniform velocity along the axis of the solenoid

Answer: 4. The electron will continue to move with uniform velocity along the axis of the solenoid

F = evBsinθ

As θ= 0, F = 0.

Important Definitions in Electromagnetism

Question 38. Consider a wire carrying a steady current I placed in a uniform magnetic field \(\vec{B}\). Consider the charges inside the wire. It is known that magnetic forces do no work. This implies that

1. The motion of charges inside the wire moves to the surface as a result of \(\vec{B}\).
2. Some charges inside the wire move to the surface as a result of \(\vec{B}\)
3. If the wire moves under the influence of \(\vec{B}\), no work is done by the force
4. If the wire moves under the influence of \(\vec{B}\), no work is done by the magnetic force on the ions, assumed to be fixed within the wire.

Answer:

2. Some charges inside the wire move to the surface as a result of \(\vec{B}\)

4. If the wire moves under the influence of \(\vec{B}\), no work is done by the magnetic force on the ions, assumed to be fixed within the wire.

⇒ \(\vec{F}=q(\vec{v} \times \vec{B}),\)

⇒ \(\vec{F}=I(\vec{l} \times \vec{B})\)

The direction of force due to the magnetic field is perpendicular to the displacement of the charge.

Question 39. A cubical region of space is filled with some uniform electric and magnetic fields. An electron enters the cube across one of its faces with velocity \(\vec{v}\) and a position enters via the opposite face with velocity \(\vec{v}\). At this instant,

1. The electric forces on both particles cause identical accelerations
2. The magnetic forces on both particles cause equal acceleration
3. Both particles gain or lose energy at the same rate
4. The motion of the center of mass is determined by \(\vec{B}\) alone

Answer:

2. The magnetic forces on both the particles cause equal acceleration

3. Both particles gain or lose energy at the same rate

4. The motion of the center of mass is determined by \(\vec{B}\) alone

For electric field, \(\vec{F}=q \vec{E}\)

So, \(\vec{F}_e=-e \vec{E} \text { and } \vec{F}_p=+e \vec{E}\)

Resultant electric force \(\vec{F}_e+\vec{F}_p=\overrightarrow{0}\)

For magnetic field, \(\vec{F}_e=-e \vec{v} \times \vec{B}\)

⇒ \(\vec{F}_p=+e(-\vec{v} \times \vec{B})=-e \vec{v} \times \vec{B}=\vec{F}_e\)

Class 12 physics electric energy questions

Question 40. A charged particle would continue to move with a constant velocity in a region where

1. \(\vec{E}=\overrightarrow{0}, \vec{B} \neq \overrightarrow{0}\)
2. \(\vec{E} \neq \overrightarrow{0}, \vec{B} \neq \overrightarrow{0}\)
3. \(\vec{E} \neq \overrightarrow{0}, \vec{B}=\overrightarrow{0}\)
4. \(\vec{E}=\overrightarrow{0}, \vec{B}=\overrightarrow{0}\)

Answer:

1. \(\vec{E}=\overrightarrow{0}, \vec{B} \neq \overrightarrow{0}\)

2. \(\vec{E} \neq \overrightarrow{0}, \vec{B} \neq \overrightarrow{0}\)

4. \(\vec{E}=\overrightarrow{0}, \vec{B}=\overrightarrow{0}\)

No force acts; if \(\vec{v} \text { and } \vec{B}\) are parallel.

When \(\vec{v}, \vec{E} \text { and } \vec{B}\) are mutually perpendicular, \(\vec{v}\) does not change if \(\frac{E}{B}=v\).

For force \(q \vec{E}(\neq 0), \vec{v}\) must change.

No force acts; so \(\vec{v}\) does not change.

Question 41. Two identical current-carrying coaxial loops, carry current I in an opposite sense. A simple campervan loop passes through both of them once. Calling the loop as C,

1. \(\oint_C \vec{B} \cdot d \vec{l}=\mp 2 \mu_0 I\)
2. The value of \(\int_C \vec{B} \cdot d \vec{l} \text { is }\) independent of the shape of C
3. There may be a point on C where \(\vec{B} \text { and } d \vec{l}\) is perpendicular
4. \(\vec{B}\) vanishes everywhere on C

Answer:

2. The value of \(\int_C \vec{B} \cdot d \vec{l} \text { is }\) independent of the shape of C

3. There may be a point on C where \(\vec{B} \text { and } d \vec{l}\) is perpendicular

⇒ \(\oint_C \vec{B} \cdot d \vec{l}=\mu_0(I-I)=0\) \(\left[∵ \oint_C \vec{B} \cdot d \vec{l}=\mu_0 I\right]\)

At the two extremities of C, \(\vec{B}\) and \(\vec{dl}\) are perpendicular to each other.

Electromagnetism MCQs for Class 12 Physics WBCHSE 

Question 42. A magnetic needle facing north-south can rotate freely in a horizontal plane. A conducting wire is placed parallel to it in a north-south direction.

1. The direction of the current is from south to north and the conductor is above the magnet-north pole of the magnet will be deflected towards west.
2. The direction of the current is from north to south, the conductor is above the magnet-south pole of the magnet will be deflected towards the west.
3. The direction of the current is from south to north, the conductor is below the magnet-the south pole of the magnet will be deflected towards west.
4. The direction of the current is from north to south, the conductor is below the magnet-the north pole of the magnet will be deflected towards west.

Answer:

1. The direction of the current is from south to north and the conductor is above the magnet-north pole of the magnet will be deflected towards west.

2. The direction of the current is from north to south, the conductor is above the magnet-south pole of the magnet will be deflected towards the west.

3. The direction of the current is from south to north, the conductor is below the magnet south pole of the magnet will be deflected towards the west.

4. The direction of the current is from north to south, the conductor is below the magnet north pole of the magnet will be deflected towards the west.

Class 11 Physics	Class 12 Maths	Class 11 Chemistry
NEET Foundation	Class 12 Physics	NEET Physics

Question 43. The magnetic field developed at a point near a straight-carrying conductor depends on

1. Material of the conductor
2. Distance of the point from the conductor
3. Direction and magnitude of current
4. Medium between the point and conductor

Answer:

2. Distance of the point from the conductor

3. Direction and magnitude of current

4. Medium between the point and conductor

Examples of Electromagnetic Induction Questions

Question 44. A current I passes through the conductor ABCD, The magnetic field at 0 In the same plane is

1. Zero For the currents in segments AD and CD
2. Zero for current in segment BC
3. \(\frac{\mu_0}{4 \pi} \cdot \frac{2 I}{r}\) for the currents in segments AB and CD
4. \(\frac{\mu_0 I \theta}{4 \pi r}\) for current in segment BC

Answer:

1. Zero For the currents in segments AD and CD

4. \(\frac{\mu_0 I \theta}{4 \pi r}\) for current in segment BC

Electromagnetism MCQs for Class 12 Physics WBCHSE 

Question 45. Two identical charged particles enter a uniform magnetic field with the same speed but at angles 30° and 60° with the field. Let a, b, and c be the ratios of their time periods, radii, and pitches of the helical paths. Then

1. abc = 1
2. abc > 1
3. abc < 1
4. a = bc

Answer:

1. abc = 1

4. a = bc

Class 12 physics electric energy questions

Question 46. In case of a current carrying solenoid

1. Internal lines of force are parallel
2. The lines of force become congested upon increasing the current
3. No north-south pole is produced in the absence of a core of magnetic material
4. The magnetic field increases on increasing the number of turns

Answer:

1. Internal lines of force are parallel

2. The lines of force become congested upon increasing the current

4. The magnetic field increases on increasing the number of turns

Question 47. A particle with charge q is moving with a velocity \(\vec{v}\) in a magnetic field \(\vec{B}\). If the force acting on the particle is \(\vec{F}\) then,

1. \(\vec{F}=0 \text {, when } \vec{v} \text { and } \vec{B}\) are parallel
2. Magnitude of \(\vec{F}\) is maximum when \(\vec{v} \text { and } \vec{B}\) are perpendicular to each other
3. \(|\vec{F}|=\frac{1}{2} q v B\) when the angle between \(\vec{v} \text { and } \vec{B}\) is 45°
4. The direction of \(\vec{F}\) remains the same when the directions of both \(\vec{v} \text { and } \vec{B}\) are reversed simultaneously

Answer:

1. \(\vec{F}=0 \text {, when } \vec{v} \text { and } \vec{B}\) are parallel

2. Magnitude of \(\vec{F}\) is maximum when \(\vec{v} \text { and } \vec{B}\) are perpendicular to each other

4. Direction of \(\vec{F}\) remains the same when the directions of both \(\vec{v} \text { and } \vec{B}\) are reversed simultaneously

Question 48. The rotation of Barlow’s wheel is due to the action of the magnetic field on the current. In the case of this wheel.

1. The rotational speed does not increase if both the current and the magnetic field are increased simultaneously
2. The rotational speed increases when the current is increased
3. The rotational speed increases when the magnetic field is increased
4. The rotational speed becomes higher as the wheel is made lighter

Answer:

2. The rotational speed increases when the current is increased

3. The rotational speed increases when the magnetic field is increased

4 The rotational speed becomes higher as the wheel is made lighter

Question 49. Two loan straight conducting wires arc kept parallel to one other at a distance r apart. When a current I passes through both the wires In the same direction, a force of attraction F acts between the wires. Which of the following statements is/are true?

1. When r – 0.5 m and I = 1 A, F = 10^-7 N
2. When r = 2.0 m and I = 1 A , F = 10^-7 N
3. When r = 8.0 m and I = 2 A, F = 10^-7 N
4. When r = 1.0 m and I = 1 A, F = 2 x 10^-7 N

Answer:

1. When r – 0.5 m and I = 1 A, F = 10^-7 N

2. When r = 2.0 m and I = 1 A , F = 10^-7 N

4. When r = 1.0 m and I = 1 A, F = 2 x 10^-7 N

Electric power multiple-choice questions

Question 50. Equal and opposite currents flow through two long straight and parallel wires. A’, B, and C are three points on the same plane.

1. The magnetic field developed at B is zero
2. The magnetic field developed at B is maximum and acts downward with respect to the plane
3. The magnetic field developed at A acts upwards with respect to the plane
4. The magnetic field developed at C acts downwards with respect to the plane

Answer:

2. The magnetic field developed at B is maximum and acts downward with respect to the plane

3. Magnetic field developed at A acts upwards with respect to the plane

Electromagnetism MCQs for Class 12 Physics WBCHSE 

Question 51. Usually, the resistance G of a galvanometer is of moderate value. But it has a full-scale deflection for a very small current I_G. In the state of full-scale deflection the terminal potential difference of the galvanometer V_G = G.I_G. Usually this V_G is also very small. If the current in an electrical circuit I > IG and the galvanometer is directly connected in series with the circuit, it cannot measure the current through the circuit; moreover, there is the possibility of the galvanometer being damaged as a result of high current. Now if a resistor of small resistance S is connected in parallel with the galvanometer as a shunt, a major portion of the current flows through the shunt. If the galvanometer shows full deflection for a current I (I > I_G) in the main circuit, then it can be said that the galvanometer is capable of measuring a current up to I i.e., it will become an ammeter of range up to a current I. On the other hand, if the potential difference between two points of the main circuit V > V_G and the galvanometer is connected in parallel with the circuit it cannot measure the potential difference V; moreover since V > V_G, the galvanometer may be damaged. Now if a resistor of high resistance R is connected in series with the galvanometer, the potential difference between the two terminals A and B of the combination is much larger than VG. If this combination shows full-scale deflection for a potential difference V (V > V_G) when connected in parallel to two points of the main circuit, then it can be said that the galvanometer is capable of measuring a potential difference up to V, i.e., it has now become a voltmeter of range up to V.

1. The resistance of the shunt to be connected with the galvanometer to convert it to an ammeter capable of measuring a current of nlG (n > 1) is

1. nG
2. \(\frac{G}{n+1}\)
3. \(\frac{G}{n}\)
4. \(\frac{G}{n-1}\)

Answer: 4. \(\frac{G}{n-1}\)

Electric power multiple-choice questions

2. The resistance to be connected in series with this galvanometer to convert it to a voltmeter capable of measuring a potential difference of nVQ (n > l) is

1. nG
2. (n+1)G
3. (n-1)G
4. \(\frac{n}{n-1} G\)

Answer: 3. (n-1)G

3. The range of the ammeter was designed by connecting a 10a shunt with a 100Ω – 100μ A galvanometer is

1. 0-1 mA
2. 0-1.1 mA
3. 0-10 mA
4. 0-11 mA

Answer: 2. 0-1.1 mA

4. A galvanometer of 500Ω is converted to an ammeter of range 0- 10 A when a 5Ω shunt is connected to it. The maximum current that can flow through the galvanometer is

1. 98 mA
2. 99 mA
3. 100 mA
4. 101 mA

Answer: 2. 99 mA

5. A 10Ω galvanometer shows full-scale deflection for a current of 10 mA. What resistance should be connected with it in series so as to convert it to a voltmeter of range 0- 5 V.

1. 490Ω
2. 499Ω
3. 500Ω
4. 510Ω

Answer: 490Ω

Electromagnetism MCQs for Class 12 Physics WBCHSE 

6. The resistance of a voltmeter is 300Ω. It shows full-scale deflection for a terminal potential difference of 150 V. What resistance should be connected in parallel with the voltmeter to convert it to an ammeter of range θ = 8 A?

1. 19Ω
2. 20Ω
3. 21Ω
4. 190Ω

Answer: 2. 20Ω

7. When a resistance R is connected in parallel to a galvanometer, the range of current measurable by it becomes n times. On the other hand, if the same resistance is connected in series, the range of potential difference measurable by it becomes n’ times. The value of n’ is

1. n-1
2. n+1
3. \(\frac{n+1}{n}\)
4. \(\frac{n}{n-1}\)

Answer: 4. \(\frac{n}{n-1}\)

Question 52. The direction of the magnetic field developed at a point near a long straight current-carrying wire can be determined by Mazwell’s cork screw rule. This rule stated that if a right-handed corkscrew is driven along the direction of current in a conductor, the direction of rotation of the thumb at any point near the conductor gives the direction of the magnetic field at that point. If the current through such a conductor is I, then the value of magnetic field developed at a distance r from the conductor in any directionis given by \(B=\frac{\mu_0 I}{2 \pi r}\), where \(\mu_0=4 \pi \times 10^{-7} \mathrm{H} \cdot \mathrm{m}^{-1}\) and unit of B is tesla (T). Let us now consider two long straight conducting wires 1 and 2 kept rigidly at a distance of 4 cm and parallel to each other. A current of 10 A flows through each wire in the same direction. In P and Q are two points on the plane of the wires and both are 1 cm away from the first wire.

1. The direction of the resultant magnetic field at P is

1. Downward with respect to the page
2. Upward with respect to the page
3. Toward left
4. Toward right

Answer: 2. Upward with respect to the page

2. Value of resultant magnetic field at P (in T)

1. 1.33 x 10^-4
2. 1.6 x 10^-4
3. 2.4 x 10^-4
4. 2.67 x 10^-4

Answer: 3. 2.4 x 10^-4

WBCHSE physics electric energy MCQs 

3. Direction of the resultant magnetic field at Q

1. Downward with respect to the page
2. Upward with respect to the page
3. Toward right
4. Toward left

Answer: 1. Downward with respect to the page

4. Value of resultant magnetic field at Q (in T)

1. 1.33 x 10^-4
2. 1.6 x 10^-4
3. 2.4 x 10^-4
4. 2.67 x 10^-4

Answer: 1. 1.33 x 10^-4

5. The direction of current is reversed in conductor 2, i.e., the current in the two wires are now in opposite directions, then the resultant magnetic field at P is

1. Downward with respect to the page
2. Upward with respect to the page
3. Toward left
4. Toward right

Answer: 2. Upward with respect to the page

Electromagnetism MCQs for Class 12 Physics WBCHSE 

6. In condition (v) the value of the resultant magnetic field at P (in T)

1. 1.33 x 10^-4
2. 1.6 x 10^-4
3. 2.4 x 10^-4
4. 2.67 x 10^-4

Answer: 2. 1.6 x 10^-4

7. In condition (v) the direction of the resultant magnetic field at Q is

1. Downward with respect to the page
2. Upward with respect to the page
3. Toward left
4. Toward right

Answer: 1. Downward with respect to the page

Question 53. A circular loop of radius a and two long parallel wires (marked I and II) are all in the plane of the paper. The distance of each wire from the center of the loop is d. The loop and the wires are carrying the same current I. The current in the loop is in the counterclockwise direction if seen from above.

1. When d≈a, but the wires are not touching the loop it is found that the net magnetic field on the axis of the loop is zero at a height h above the loop. In that case

1. Current in wire I and wire II is in the direction PQ and RS, respectively, and h ≈ a
2. Current in wire I and wire II is in the direction PQ and SR, respectively, and h ≈ a
3. Current in wire I and wire II is in the direction PQ and SR respectively and h≈1.2a
4. Current in wire I and wire II is in the direction PQ and RS respectively and h ≈ 1.2a

Answer: 3. Current in wire I and wire II is in the direction PQ and SR respectively and h≈1.2a

Electromagnetism MCQs for Class 12 Physics WBCHSE 

2. Consider d>>a and the loop is rotated about its diameter parallel to the wires by 30° from the position. If the current in the wires is in the opposite directions, the torque on the loop at its new position will be (consider that the net field due to the wires is constant over the loop)

1. \(\frac{\mu_0 I^2 a^2}{d}\)
2. \(\frac{\mu_0 I^2 a^2}{2 d}\)
3. \(\frac{\sqrt{3} \mu_0 1^2 a^2}{d}\)
4. \(\frac{\sqrt{3} \mu_0 1^2 a^2}{2 d}\)

Answer: 2. \(\frac{\mu_0 I^2 a^2}{2 d}\)

Question 54. There is a constant homogeneous electric field of 100 V.m^-1 within the region x = 0 and x = 0.167 m along the positive x-direction. There is a constant homogeneous magnetic field within the region x = 0.167 m and x = 0.334 m along the z-direction. A proton at rest at the origin (x = 0, y = 0) is released along the positive x-direction.

1. Acceleration produced in the proton is (in m.s^-2)

1. 9.58 x 10⁹ \(\hat{i}\)
2. 5.98 x 10⁹ \(\hat{i}\)
3. 8.95 x 10⁹ \(\hat{i}\)
4. 3.33 x 10⁸ \(\hat{i}\)

Answer: 9.58 x 10⁹ \(\hat{i}\)

WBCHSE physics electric energy MCQs 

2. The path described by the proton in the magnetic field is

1. Straight
2. Circular
3. Parabolic
4. Elliptical

Answer: 2. Circular

3. The minimum value of the magnetic field (\(\vec{B}\)) necessary to confirm the proton within the region x = 0.167 m to x = 0.334 m is

1. 5 X 10^-1 T
2. 8 X 10^–4 T
3. 6 X 10^-2 T
4. 7 X 10^-3 T

Answer: 4. 7 X 10^-3 T

Question 55. A proton with a speed of 2 x 10⁷ m.s^-1 enters a magnetic field of flux density 1.5 Wb.m^-2, making an angle of 30° with the field. The force acting on the proton is

1. 2.4 x 10^-14 N
2. 0.24 x 10^-12 N
3. 0.024 x 10^-24 N
4. 24 x 10^-12 N

Answer: 1. 2.4 x 10^-14 N

We know, \(|F|=q v B \sin \theta\)

given, B = 1.5 Wb.m^-2, v = 2 x 10⁷ m.s^-1, 6 = 30°

q = 1.6 x 10^-19 C

Then, \(F=1.6 \times 10^{-19} \times 2 \times 10^7 \times 1.5 \times \frac{1}{2}=2.4 \times 10^{-12} \mathrm{~N}\)

Electromagnetism MCQs for Class 12 Physics WBCHSE 

Question 56. A straight conductor of length l m carrying a current IA is bent in the form of a semicircle. The magnetic field (in tesla) at the center of the semicircle is

1. \(\frac{\pi^2 I}{l} \times 10^{-7}\)
2. \(\frac{\pi I}{l} \times 10^{-7}\)
3. \(\frac{\pi I}{l^2} \times 10^{-7}\)
4. \(\frac{\pi R^2}{l} \times 10^{-7}\)

Answer: 1. \(\frac{\pi^2 I}{l} \times 10^{-7}\)

Ifradius of the semicircle is r,

then, \(\pi r=l \quad \text { or, } r=\frac{l}{\pi}\)

Now, the magnetic field at the center of the semicircle,

⇒ \(B=\frac{\mu_0 I}{4 r}=\frac{\mu_0 I \pi}{4 l}=\frac{\mu_0}{4 \pi} \cdot \frac{\pi^2 I}{l}\)

⇒ \(\frac{\pi^2 I}{l} \times 10^{-7}\left[∵ \frac{\mu_0}{4 \pi}=10^{-7}\right]\)

The option 1 is correct

Question 57. A galvanometer having Internal resistance 10Ω requires 0.01 A for a full-scale deflection. To convert this galvanometer to a voltmeter of full-scale deflection at 120 V, we need to connect a resistance of

1. 11990Ω in series
2. 11990Ω in parallel
3. 12010Ω in series
4. 12010Ω in parallel

Answer: 1. 11990Ω in series

The internal resistance of the galvanometer (G) is 10Ω. A high-resistance R is to be added in series with a galvanometer. In this case, if the galvanometer is connected between the two points A and B then the galvanometer will become a voltmeter of voltage range from 0 to (V_A – V_B).

V_A – V_B = I_G(G+ R)

∴ \(R=\frac{V_A-V_B}{I_G}-G\)

⇒ \(\frac{120}{0.01}-10\)

=12000-10

= 11990Ω

The option 1 is correct.

Electromagnetism MCQs for Class 12 Physics WBCHSE 

Question 58. A long conducting wire carrying a current I is bent at 120°. The magnetic field B at a point on the right bisector of the bending angle at a distance d from the bend is (μ₀ is the permeability of free space)

1. \(\frac{3 \mu_0 I}{2 \pi d}\)
2. \(\frac{\mu_0 I}{2 \pi d}\)
3. \(\frac{\mu_0 I}{\sqrt{3} \pi d}\)
4. \(\frac{\sqrt{3} \mu_0 I}{2 \pi d}\)

Answer: 4. \(\frac{\sqrt{3} \mu_0 I}{2 \pi d}\)

⇒ \(P Q=d, P R=\frac{d \sqrt{3}}{2}=P S\)

The magnetic field at P due to the portion QT,

⇒ \(B=\frac{\mu_0}{4 \pi} \cdot \frac{I}{\frac{d \sqrt{3}}{2}}\left(\sin 90^{\circ}+\sin 30^{\circ}\right)\)

⇒ \(\frac{\mu_0}{4 \pi} \cdot \frac{2 I}{d \sqrt{3}} \cdot \frac{3}{2}=\frac{\sqrt{3} \mu_0 I}{4 \pi d}\)

Magnetic field at P due to the portion \(Q U, B_2=\frac{\sqrt{3} \mu_0 I}{4 \pi d}\)

∴ Total magnetic field at \(P=B_1+B_2=\frac{\sqrt{3} \mu_0 I}{2 \pi d}\)

The option 4 is correct

WBCHSE physics electric energy MCQs 

Question 59. A proton of mass m and charge q is moving in a plane with kinetic energy E. If there exists a uniform magnetic field B, perpendicular to the plane of the motion, the proton will move in a circular path of radius

1. \(\frac{2 E m}{q B}\)
2. \(\frac{\sqrt{2 E m}}{q B}\)
3. \(\frac{\sqrt{E m}}{2 q B}\)
4. \(\sqrt{\frac{2 E q}{m B}}\)

Answer: 2. \(\frac{\sqrt{2 E m}}{q B}\)

The kinetic energy of the proton,

⇒ \(E=\frac{1}{2} m v^2=\frac{1}{2} m \frac{q^2 B^2 r^2}{m^2}\)

⇒ \(\frac{q^2 B^2 r^2}{2 m}\left[∵ \text { Velocity of the proton, } v=\frac{q B r}{m}\right]\)

∴ \(r^2=\frac{2 m E}{q^2 B^2} \quad \text { or, } r=\frac{\sqrt{2 E m}}{q B}\)

The option 2 is correct.

Electromagnetism MCQs for Class 12 Physics WBCHSE 

Question 60. A stream of electrons and protons are directed towards a narrow slit on a screen. The intervening region has a uniform electric field \(\vec{E}\) (vertically downwards) and a uniform magnetic field \(\vec{B}\).

1. Electrons and proton with speed \(\frac{|\vec{E}|}{|\vec{B}|}\) will pass through the slit
2. Protons with speed \(\frac{|\vec{E}|}{|\vec{B}|}\) will pass through the slit, electrons of the same speed will not
3. Neither electrons nor proton will go through the slit irrespective of their speed
4. Electrons will always be deflected upwards irrespective of their speed

Answer: 3 and 4 are correct

Let the directions of \(\vec{E}, \vec{v} \text { and } \vec{B}\) along the x-axis, y-axis, and z-axis respectively.

The electric force +eE and magnetic force +evB both act on the proton along the x-axis. So the proton will be deflected from the y-axis. Similarly, as the electron is negatively charged the two forces will deflect the electron along the negative x-axis. So neither the electron nor the proton will go through the slit.

The options 3 and 4 are correct.

Question 61. Two particles A and B, having equal charges, after being accelerated through the same potential difference enter a region of uniform magnetic field and the particles describe circular paths of radii R₁ and R₂ respectively. The ratio of the masses of A and B is

1. \(\sqrt{R_1 / R_2}\)
2. \(R_1 / R_2\)
3. \(\left(R_1 / R_2\right)^2\)
4. \(\left(R_2 / R_1\right)^2\)

Answer: 3. \(\left(R_1 / R_2\right)^2\)

For a circular path, magnetic force = centripetal force

i.e., \(q v B=\frac{m v^2}{R}\)

again, \(\frac{1}{2} m v^2=q V\)

hence, \(q v B=\frac{2 q V}{R}\)

or, \(R=\frac{2 V}{v B}=\frac{2 V}{B} \sqrt{\frac{m}{2 q V}}\left[∵ v=\sqrt{\frac{2 q V}{m}}\right]\)

For a circular path, magnetic force = centripetal force.

Therefore, \(\frac{R_1}{R_2}=\sqrt{\frac{m_1}{m_2}} \text { or, } \frac{m_1}{m_2}=\left(\frac{R_1}{R_2}\right)^2\)

The option 3 is correct.

Question 62. A rectangular coil carrying current is placed in a nonuniform magnetic field. On that coil the total

1. Force is non-zero
2. Force is zero
3. Torque is zero
4. Torque is non-zero

Answer: 1 and 4 are correct

If a current-carrying coil is placed in a non-uniform magnetic field, then, in general, both net force and net torque are non-zero. The options 1 and 4 are correct.

Electromagnetism MCQs for Class 12 Physics WBCHSE 

Question 63. The magnetic field due to a current in a straight wire segment of length L at a point on its perpendicular bisector at a distance r(r>>L)

1. \(\text { decreases as } \frac{1}{r}\)
2. \(\text { decreases as } \frac{1}{r^2}\)
3. \(\text { decreases as } \frac{1}{r^3}\)
4. \(\text { approaches a finite limit as } r \rightarrow \infty\)

Answer: 2. \(\text { decreases as } \frac{1}{r^2}\)

According to Biot-Savart law, \(d B \propto \frac{1}{r^2}\)

The option 2 is correct.

WBCHSE physics electric energy MCQs 

Question 64. The magnets of two suspended coil galvanometers are of the same strength so that they produce identical uniform magnetic fields in the region of the coils. The coil of the first one is in the shape of a square of side a and that of the second one is circular of radius \(\frac{a}{\sqrt{\pi}}\). When the same current is passed through the coils, the ratio of the torque experienced by the first coil to that experienced by the second one is

1. \(1: \frac{1}{\sqrt{\pi}}\)
2. 1:1
3. π: 1
4. 1: π

Answer: 2. 1:1

In this case, applied torque (\(\tau\)) does not depend on the shape of the coil but on the area of cross-section (A) of the coil.

∴ \(\frac{\tau_1}{\tau_2}=\frac{A_1}{A_2}=\frac{a^2}{\pi\left(\frac{a}{\sqrt{\pi}}\right)^2}=1\)

or, \(\tau_1: \tau_2=1: 1\)

The option 2 is correct.

Question 65. A proton is moving with a uniform velocity of 10⁶ m.s^-1 along the F-axis, under the joint action of a magnetic field along the z-axis and an electric field of magnitude 2 x 10⁴ V.m^-1 along the negative X-axis. If the electric field is switched off, the proton starts moving in a circle. The radius of the circle is nearly (given: \(\frac{e}{m}\)ratio for proton ≈ 108 C.kg^-1 )

1. 0.5 m
2. 0.2 m
3. 0.1m
4. 0.05 m

Answer: 1. 0.5 m

Initially, the Lorentz force acting on the proton,

⇒ \(\vec{F}=\vec{F}_e+\vec{F}_m=\vec{E} q+q \vec{v} \times \vec{B}=-E q \hat{i}+q v B \hat{i}\)

Since the proton has no acceleration,

∴ \(\vec{F}=0 \quad ∴ E q=q v B\)

or, \(B=\frac{E}{v}=\frac{2 \times 10^4}{10^6}=2 \times 10^{-2} \mathrm{~T}\)

When the electric field is switched off the radius of the circular path,

⇒ \(R=\frac{m v}{q B}=\frac{10^6}{10^8 \times 2 \times 10^{-2}}=0.5 \mathrm{~m}\)

The option 1 is correct.

Question 66. Two long parallel wires separated by 0.1 m carry currents of 1A and 2A respectively in opposite directions. A third current-carrying wire parallel to both of them is placed in the same plane such that it feels no net magnetic force. It is placed at a distance of

1. 0.5 m from the 1st wire, towards the 2nd wire
2. 0.2 m from the 1st wire, towards the 2nd wire
3. 0.1 m from the 1st wire, away from the 2nd wire
4. 0.2 m from the 1st wire, away from the 2nd wire

Answer: 3. 0.1 m from the 1st wire, away from the 2nd wire

Let the 3rd wire be at a distance x from the first wire.

Since, \(B_1=B_2 \quad \text { or, } \frac{2 i_1}{x}=\frac{2 i_2}{(0.1+x)} \quad \text { or, } x=0.1 \mathrm{~m}\)

The option 3 is correct.

Real-Life Applications of Electromagnetic Principles

Question 67. A proton of mass m moving with a speed V << c, velocity of light in vacuum) completes a circular orbit in time T in a uniform magnetic field. If the speed of the proton is increased to √2v, what will be the time needed to complete the circular orbit?

1. √2
2. T
3. \(\frac{T}{\sqrt{2}}\)
4. \(\frac{T}{2}\)

Answer:

The magnetic force on a proton of mass m moving along a circular path in a uniform magnetic field,

⇒ \(\vec{F}_m=q(\vec{v} \times \vec{B})\)

∴ \(F_m=q v B \sin 90^{\circ}\) [B = magnetic field, q = charge of proton]

= qvB

Since the proton is moving along a circular path of radius r,

⇒ \(\frac{m v^2}{r}=q v B \quad .. v=\frac{q B r}{m}\)

Now the time taken by the proton to make one complete revolution is \(\frac{2 \pi r}{v}\)

Hence, \(T=\frac{2 \pi r}{v} \text { or, } T=\frac{2 \pi r \times m}{q B r} \text { or, } T=\frac{2 \pi m}{q B}\)

Since T does not depend on the velocity ofdie proton, if the velocity changes to V2v then T will remain unchanged.

The option 2 is correct.

WBCHSE Class 12 Physics Electromagnetic Theory MCQs 

Question 68. A uniform current is flowing along the length of an infinite, straight, thin, hollow cylinder of radius R. The magnetic field B produced at a perpendicular distance d from the axis of the cylinder is plotted in a graph. Which of the following figures looks like the plot?

Answer: 3.

Applying Ampere’s circuital law, a magnetic field at a point outside the cylinder and located at a perpendicular distance d from its axis,

⇒ \(B=\frac{\mu_0 I}{2 \pi d} \text { if, } d>R\)

⇒ \(\text { If } d=R \text { then, } B=\frac{\mu_0 I}{2 \pi R}\)

If d < R, B = 0 [since no current flows inside a hollow cylinder]

∴ \(B \propto \frac{1}{d}\)

The option 3 is correct.

WBCHSE physics electric energy MCQs 

Question 69. A circular loop of radius r of conducting wire connected with a voltage source of zero internal resistance produces a magnetic field B at its center. If instead, a circular loop of radius 2r, made of the same material, having the same cross-section is connected to the same voltage source, what will be the magnetic field at its center?

1. \(\frac{B}{2}\)
2. \(\frac{B}{4}\)
3. 2B
4. B

Answer: 2. \(\frac{B}{4}\)

The magnetic field at the center of a circular conducting loop,

⇒ \(B=\frac{\mu_0 I}{2 r}\) [I = current in the loop and r = radius of the loop]

If the resistance of the loop in the first case is R, then the
resistance in the second case,

R’ = 2R [since in the second case the length of the loop is doubled]

∴ In the second case, current in the loop, \(I^{\prime}=\frac{I}{2}\)

∴ \(B^{\prime}=\frac{\mu_0 I^{\prime}}{2 \times 2 r}=\frac{\mu_0 \times \frac{I}{2}}{2 r \times 2}=\frac{B}{4}\)

The option 2 is correct.

Question 70. A light-charged particle is revolving in a circle of radius r in the electrostatic attraction of a static heavy particle with an opposite charge. How does the magnetic field B at the center of the circle due to the moving charge depend on r?

1. \(B \propto \frac{1}{r}\)
2. \(B \propto \frac{1}{r^2}\)
3. \(B \propto \frac{1}{r^{3 / 2}}\)
4. \(B \propto \frac{1}{r^{5 / 2}}\)

Answer: 4. \(B \propto \frac{1}{r^{5 / 2}}\)

Let a light particle of mass m and charge -q2 revolve around a heavy particle of charge +q1 in a circular path of radius r with a velocity v

Now, \(\frac{m v^2}{r}=\frac{1}{4 \pi \epsilon_0} \frac{\left(+q_1\right) \times\left(-q_2\right)}{r^2}\)

or, \(v^2=\frac{1}{4 \pi \epsilon_0} \cdot \frac{-q_1 q_2}{m r}\)

∴ \(v \propto \frac{1}{\sqrt{r}}\)

The time period of the charge in a circular orbit,

⇒ \(T=\frac{2 \pi r}{v}\)

∴ Currently due to the revolving charge,

⇒ \(I=\frac{-q_2}{T}=\frac{q_2 \times v}{2 \pi r}\)

∴ \(I \propto \frac{v}{r} \quad \text { or, } I \propto \frac{1}{r^{3 / 2}}\left[∵ \nu \propto \frac{1}{\sqrt{r}}\right]\)

So, the magnetic field at the center of the circle due to the moving charge +q2,

⇒ \(B=\frac{\mu_0 I}{2 r} \quad \text { or, } B \propto \frac{I}{r}\)

or, \(B \propto \frac{1}{r^{3 / 2} \times r} \quad \text { or, } B \propto \frac{1}{r^{5 / 2}}\)

The option 4 is correct.

WBCHSE Class 12 Physics Electromagnetic Theory MCQs 

Question 71. A conductor lies along the z-axis at -1.5 < z < 1.5 m and carries a fixed current of 10.0 A in the -ve z-direction. For a field \(B=3.0 \times 10^{-4} e^{-0.2 x} \hat{a}_y\), find the power required to move the conductor at a constant speed to r = 2.0m, y=0 in 5 x 10-3s. Assume parallel motion along the x-axis. Here \(\hat{a}_x, \hat{a}_y, \hat{a}_z\) are unit vectors along the x, y, and z axes, respectively.

1. 1.57 W
2. 2.97 W
3. 14.85 W
4. 29.7 W

Answer: 2. 2.97 W

Work, \(W=\int_0^2 F d x\)

F = BIl

⇒ \(3 \times 10^{-4} e^{-0.2 x} \times 10 \times 3[∵ l=1.5+1.5=3 \mathrm{~m}]\)

∴ \(W=\int_0^2 3 \times 10^{-4} e^{-0.2 x} \times 10 \times 3 d x\)

⇒ \(9 \times 10^{-3} \int_0^2 e^{-0.2 x} d x\)

⇒ \(=\frac{9 \times 10^{-3}}{0.2}\left(-e^{-0.2 \times 2}+1\right)=\frac{2.97 \times 10^{-3}}{0.2} \mathrm{~J}\)

Power, \(P=\frac{W}{t}=\frac{2.97 \times 10^{-3}}{0.2 \times 5 \times 10^{-3}}=2.97 \mathrm{~W}\)

The option 2 is correct.

Question 72. Two long current-carrying thin wires, both with current I, are held by insulating threads of length L and are in equilibrium as shown in the figure, with threads making an angle 0 with the vertical. If the wires have mass A, per unit length then the value of I is (g = gravitational acceleration)

1. \(\sin \theta \sqrt{\frac{\pi \lambda g L}{\mu_0 \cos \theta}}\)
2. \(2 \sin \theta \sqrt{\frac{\pi \lambda g L}{\mu_0 \cos \theta}}\)
3. \(2 \sqrt{\frac{\pi g L}{\mu_0} \tan \theta}\)
4. \(\sqrt{\frac{\pi \lambda g L}{\mu_0} \tan \theta}\)

Answer: 2. \(2 \sin \theta \sqrt{\frac{\pi \lambda g L}{\mu_0 \cos \theta}}\)

The current I is flowing in opposite directions in the two wires.

Therefore, the repulsive force per unit length of the wires,

⇒ \(F=\frac{\mu_0}{4 \pi} \cdot \frac{2 I^2}{2 x}=\frac{\mu_0}{4 \pi} \frac{P^2}{L \sin \theta} \quad[x=L \sin \theta]\)

Weight of the wire per unit length = λg.

If T is the tension in the string, in equilibrium condition,

Tsinθ = F and Tcosθ = λg

⇒ \(\tan \theta=\frac{F}{\lambda g}=\frac{\mu_0 I^2}{4 \pi \lambda g L \sin \theta}\)

⇒ \(I=\sqrt{\frac{4 \pi \lambda g L \sin ^2 \theta}{\mu_0 \cos \theta}}=2 \sin \theta \sqrt{\frac{\pi \lambda g L}{\mu_0 \cos \theta}}\) \(\left[∵ \sin \theta \tan \theta=\frac{\sin ^2 \theta}{\cos \theta}\right]\)

The option 2 is correct.

WBCHSE Class 12 Physics Electromagnetic Theory MCQs 

Question 73. Two coaxial solenoids of different radii carry current I in the same direction. Let \(\vec{F}_1\) be the magnetic force on the inner solenoid due to the outer one and \(\vec{F}_2\) be the magnetic force on the outer solenoid due to the inner one. Then

1. \(\vec{F}_1=\vec{F}_2=0\)
2. \(\vec{F}_1 \text { is radially inwards and } \vec{F}_2 \text { is radially outwards }\)
3. \(\vec{F}_1 \text { is radially inwards and } \vec{F}_2=0\)
4. \(\overrightarrow{F_1} \text { is radially outwards and } \overrightarrow{F_2}=0\)

Answer: 1. \(\vec{F}_1=\vec{F}_2=0\)

The two coaxial solenoids are actually two magnets of the same strength, the inner solenoid being completely surrounded by the outer solenoid. No resultant force due to a magnet acts on another magnet.

Hence, \(\vec{F}_1=\vec{F}_2=0\)

The option 1 is correct.

Question 74. A rectangular loop of sides 10 cm and 5 cm carrying a current I of 12 A is placed in different orientations as shown in the figures below

1. Stable equilibrium
2. Unstable equilibrium

If there is a uniform magnetic field of 00.3 T in the positive z direction, in which orientations the loop would be in

1. (1) and (2), respectively
2. (1) and (3), respectively
3. (2) and (4), respectively
4. (2) and (3), respectively

Answer: 3. (2) and (4), respectively

In cases (2) and (4), torque acting on the loop due to B = 0.

In case (2), if the rectangular loop is slightly disturbed from the xy-plane, the torque acts in such a direction that it tries to bring the loop back to the xy-plane. On the other hand, in case (4), for a similar disturbance of the loop, the acting torque tends to take the loop away from the xy-plane.

It can be noted that in all the cases from (2) to (4), the resultant force acting on the loop = 0.

The option 3 is correct

Question 75. Two identical wires A and B, each of length l, carry the same current I. Wire A is bent into a circle of radius R and wire B is bent to form a square of side a. If BA and BB are the values of the magnetic field at the centers of the circle and square respectively, then the ratio B_A/B_B is

1. \(\frac{\pi^2}{8}\)
2. \(\frac{\pi^2}{16 \sqrt{2}}\)
3. \(\frac{\pi^2}{16}\)
4. \(\frac{\pi^2}{8 \sqrt{2}}\)

Answer: 4. \(\frac{\pi^2}{8 \sqrt{2}}\)

We have, \(l=2 \pi R \quad \text { or, } R=\frac{l}{2 \pi}\)

Again, \(l=4 a \quad \text { or, } a=\frac{l}{4}\)

The magnetic field at the center of the circle,

⇒ \(B_A=\frac{\mu_0 I}{2 R}=\frac{\mu_0 \pi I}{l}\)

The magnetic field at the center of the square,

⇒ \(B_B=4 \times \frac{\mu_0 I}{4 \pi \times \frac{a}{2}}\left(\sin 45^{\circ}+\sin 45^{\circ}\right)\)

⇒ \(\frac{\mu_0 I}{\frac{\pi l}{8}} \times \frac{2}{\sqrt{2}}=\frac{16 \mu_0 I}{\pi \sqrt{2} l}\)

∴ \(\frac{B_A}{B_B}=\frac{\mu_0 \pi I}{l} \times \frac{\pi \sqrt{2} l}{16 \mu_0 I}=\frac{\pi^2}{8 \sqrt{2}}\)

The option 4 is correct.

WBCHSE Class 12 Physics Electromagnetic Theory MCQs 

Question 76. When a current of 5 mA is passed through a galvanometer having a coil of resistance 15Ω, it shows full-scale deflection. The value of the resistance to be put in series with the galvanometer to convert it into a voltmeter of range 0- 10 V is

1. 1.985 x 10³Ω
2. 2.045 x 10³Ω
3. 2.535 x 10³Ω
4. 4.005 x 10³Ω

Answer: 1. 1.985 x 10³Ω

Let the value of the resistance be put in series by R.

∴ V= i_g(R_G+R)

or, \(R=\frac{V}{i_g}-R_G=\frac{10}{5 \times 10^{-3}}-15\)

= 1.985 x 10³Ω

The option 1 is correct.

Question 77. An electron, a proton, and an alpha particle having the same kinetic energy are moving in circular orbits of radii r_e, r_p, and r_α respectively in a uniform magnetic field B. The relation between r_g, r_p, r_α is

1. \(r_e<r_p<r_\alpha\)
2. \(r_e<r_\alpha<r_p\)
3. \(r_e>r_p, r_e>r_\alpha\)
4. \(r_e<r_p, r_e<r_\alpha\)

Answer: 4. \(r_e<r_p, r_e<r_\alpha\)

For a charged particle moving in a circular orbit,

⇒ \(r=\frac{\sqrt{2 m E}}{q B}\) [ r = radius of the circular orbit, E = kinetic energy of the particle]

∴ \(\frac{r_a}{r_p}=\frac{\sqrt{2 m_a}}{q_a} \times \frac{q_p}{\sqrt{2 m_p}}=1\) \(\text { since } m_\alpha=4 m_p \text { and } q_\alpha=2 q_p\)

∴ r_α = r_p

Now, since the mass of an electron is less than the mass of both proton and a -particle, r_e will be less than both r_p and r_α.

r_e< r_p , r_e < r_α

The option 4 is correct.

Question 78. Two identical long conducting wires AOB and COD are placed at right angles to each other, with one above the other such that 0 is their common point for two. The wires carry I₁ and I₂ currents respectively. A point P is at a height d above the point 0, with respect to the plane of the wires. The magnetic field at P is,

1. \(\frac{\mu_0}{2 \pi d}\left(\frac{I_1}{I_2}\right)\)
2. \(\frac{\mu_0}{2 \pi d}\left(I_1+I_2\right)\)
3. \(\frac{\mu_0}{2 \pi d}\left(I_1^2-I_2^2\right)\)
4. \(\frac{\mu_0}{2 \pi d}\left(I_1^2+I_2^2\right)^{1 / 2}\)

Answer: 4. \(\frac{\mu_0}{2 \pi d}\left(I_1^2+I_2^2\right)^{1 / 2}\)

The magnetic field at P due to current I₁,

⇒ \(B_1=\frac{\mu_0}{4 \pi} \frac{2 I_1}{d}\)

The magnetic field at P due to current I₂,

⇒ \(B_2=\frac{\mu_0}{4 \pi} \frac{2 I_2}{d}\)

∴ The resultant magnetic field due to AOB and COD,

⇒ \(B=\sqrt{\left(\frac{\mu_0}{4 \pi} \frac{2 I_1}{d}\right)^2+\left(\frac{\mu_0}{4 \pi} \frac{2 I_2}{d}\right)^2}\)

∴ \(B=\frac{2 \mu_0}{4 \pi d} \sqrt{I_1^2+I_2^2}\)

⇒ \(\frac{\mu_0}{2 \pi d}\left(I_1^2+I_2^2\right)^{\frac{1}{2}}\)

The option 4 is correct.

WBCHSE Class 12 Physics Electromagnetic Theory MCQs 

Question 79. A wire carrying a current I has the shape as shown in the adjoining figure. Linear parts of the wire are very long and parallel to the x-axis while a semicircular portion of radius R is lying on the yz-plane. The magnetic field at point O is

1. \(\vec{B}=\frac{\mu_0}{4 \pi} \frac{I}{R}(\pi \hat{i}+2 \hat{k})\)
2. \(\vec{B}=-\frac{\mu_0}{4 \pi} \frac{I}{R}(\pi \hat{i}-2 \hat{k})\)
3. \(\vec{B}=-\frac{\mu_0}{4 \pi} \frac{I}{R}(\pi \hat{i}+2 \hat{k})\)
4. \(\vec{B}=\frac{\mu_0}{4 \pi} \frac{I}{R}(\pi \hat{i}-2 \hat{k})\)

Answer: 3. \(\vec{B}=-\frac{\mu_0}{4 \pi} \frac{I}{R}(\pi \hat{i}+2 \hat{k})\)

The magnetic fields of the linear parts of the wire are directed downwards, i.e., along the negative z-direction; the value of the magnetic field in each case

⇒ \(\frac{\mu_0}{4} \frac{I}{R}\)

Again, the magnetic field for the semicircular portion of the wire at point 0 is directed along the negative x-axis and its value

⇒ \(\frac{1}{2} \frac{\mu_0 I}{2 R}=\frac{\mu_0 I}{4 R}\)

∴ Resultant magnetic field

⇒ \(-\hat{k}^{\mu_0} \frac{I}{4 \pi} \frac{I}{R}-\hat{k}^{\mu_0} \frac{I}{4 \pi} \frac{I}{R}-\hat{i} \frac{\mu_0 I}{4 R}\)

⇒ \(-\frac{\mu_0}{4 \pi} \frac{I}{R}(\pi \hat{i}+2 \hat{k})\)

The option 3 is correct.

Question 80. An electron moving in a circular orbit of radius r makes n rotations per second. The magnetic field produced at the center has a magnitude

1. \(\frac{\mu_0 n e}{2 \pi r}\)
2. Zero
3. \(\frac{\mu_0 n^2 e}{r}\)
4. \(\frac{\mu_0 n e}{2 r}\)

Answer: 4. \(\frac{\mu_0 n e}{2 r}\)

Equivalent current, \(I=\frac{e}{T}=\frac{e}{\frac{1}{n}}=e n\)

Hence, \(B=\frac{\mu_0 I}{2 r}=\frac{\mu_0 n e}{2 r}\)

The option 4 is correct.

Question 81. A square loop ABCD carrying a current i Is placed near and coplanar with a long straight conductor XY carrying a current I. The net force on the loop will be

1. \(\frac{\mu_0 I i}{2 \pi}\)
2. \(\frac{2 \mu_0 I i L}{3 \pi}\)
3. \(\frac{\mu_0 I i L}{2 \pi}\)
4. \(\frac{2 \pi_0 I i}{3 \pi}\)

Answer: 4. \(\frac{2 \pi_0 I i}{3 \pi}\)

The force of attraction on part AB of the loop by conductor XY

⇒ \(\frac{\mu_0 I i L}{2 \pi \times \frac{L}{2}}=\frac{\mu_0 I i}{\pi}\)

Similarly, the repulsive force on part CD of the loop by conductor XY

⇒ \(\frac{\mu_0 I i L}{2 \pi \times 3 \frac{L}{2}}=\frac{\mu_0 I i}{3 \pi}\)

The magnetic force on AD and BC by the conductor XY will be zero. Therefore, the net attractive force on the loop

⇒ \(A B C D=\frac{\mu_0 I i}{\pi}-\frac{\mu_0 I i}{3 \pi}=\frac{2 \pi_0 I i}{3 \pi}\)

The option 4 is correct

Question 82. A long straight wire of radius a carries a steady current I. The current is uniformly distributed over its cross-section. The ratio of the magnetic fields B and B’, at a radial distance \(\frac{a}{2}\) and 2a respectively, from the axis of the wire is

1. \(\frac{1}{2}\)
2. 1
3. 4
4. \(\frac{1}{4}\)

Answer: 2. 1

Magnetic field B at radial distance \(\frac{a}{2}\) from the axis of the wire can be calculated using Ampere’s circuital law as follows,

⇒ \(B \times 2 \pi \frac{a}{2}=\mu_0 \frac{I}{\pi a^2} \times \pi\left(\frac{a}{2}\right)^2 \quad \text { or, } B=\frac{\mu_0 I}{4 \pi a}\)

Similarly, the magnetic field B’ at radial distance 2a is given by,

⇒ \(B^{\prime} \times 2 \pi \times 2 a=\mu_0 I \quad \text { or, } B^{\prime}=\frac{\mu_0 I}{4 \pi a} \quad ∴ \frac{B}{B^{\prime}}=1\)

The option 2 is correct.

WBCHSE Class 12 Physics Electromagnetic Theory MCQs 

Question 83. An electron moves straight inside a charged parallel plate capacitor of uniform charge density σ. The space between the plates is filled with the uniform magnetic field of intensity B, as shown in the figure. Neglecting the effect of gravity, the time of straight-line motion of the electron in the capacitor is

1. \(\frac{\epsilon_0 l B}{\sigma}\)
2. \(\frac{\sigma}{\epsilon_0 l B}\)
3. \(\frac{\epsilon_0 B}{\sigma}\)
4. \(\frac{\sigma}{\epsilon_0 B}\)

Answer: 1. \(\frac{\epsilon_0 l B}{\sigma}\)

Electric field between the capacitor, \(E=\frac{\sigma}{\epsilon_0}\)

Force upon the electron = \(-e(\vec{E}+\vec{v} \times \vec{B})\)

For the straight-line motion of the electron, this force must be zero.

∴ \(\vec{E}=-\vec{v} \times \vec{B}\)

Taking only the magnitudes,

⇒ \(E=\nu B \text { or, } v=\frac{E}{B}=\frac{\sigma}{\epsilon_0 B}\)

∴ Time taken by the electron, \(t=\frac{l}{v}=\frac{\epsilon_0 l B}{\sigma}\)

The option 1 is correct.

Conceptual Questions on Ampere’s Law and Faraday’s Law

Question 84. A uniform magnetic field of 0.3 T is established along the positive Z-direction. A rectangular loop in XY-plane of sides 10 cm and 5 cm carries a current of I = 12A as shown. The torque on the loop is

1. \(+1.8 \times 10^{-2} \hat{i} \mathrm{~N} \cdot \mathrm{m}\)
2. \(-1.8 \times 10^{-2} \hat{j} \mathrm{~N} \cdot \mathrm{m}\)
3. Zero
4. \(-1.8 \times 10^{-2} \hat{i} \mathrm{~N} \cdot \mathrm{m}\)

Answer: Zero

Torque on a loop, \(\tau=I \vec{A} \times \vec{B}\)

In the given figure the area vector \(\vec{A}\) of the xy-plane is along the positive Z-direction.

also \(\vec{B}\) is directed along the positive Z-direction.

then, \(\vec{A} \times \vec{B}=\overrightarrow{0}\)

The option 3 is correct.

Question 85. A metallic rod of mass per unit length 0.5 kg m^-1 is lying horizontally on a smooth inclined plane which makes an angle of 30° with the horizontal. The rod is not allowed to slide down by flowing a current through it when a magnetic field of induction 0.25 T is acting on it in the vertical direction. The current flowing in the rod to keep it stationary is

1. 14.76 A
2. 5.98 A
3. 7.14 A
4. 11.32 A

Answer: 4. 11.32 A

For the equilibrium of the metallic rod,

mgsin30° = Bilcos30°

⇒ \(i=\frac{m g}{B l} \tan 30^{\circ}=\left(\frac{m}{l}\right) \frac{g}{B} \tan 30^{\circ}\)

⇒ \(=0.5 \times \frac{9.8}{0.25} \times \tan 30^{\circ}\)

= 11.32 A

The option 4 is correct.

WBCHSE Class 12 Physics Electromagnetic Theory MCQs 

Question 86. Current sensitivity of a moving coil galvanometer is 5 div/mA and its voltage sensitivity (angular deflection per unit voltage applied) is 20 div/V. The resistance of the galvanometer is

1. 250Ω
2. 25Ω
3. 40Ω
4. 500Ω

Answer: 1. 250Ω

The current sensitivity of the galvanometer,

⇒ \(\frac{\theta}{I}=I_s=\frac{B N A}{c}\)

Voltage sensitivity of the galvanometer,

⇒ \(\frac{\theta}{V}=V_s=\frac{B N A}{c R_G}\)

or, \(V_s=\left(\frac{B N A}{c}\right) \times \frac{1}{R_G}=I_s \times \frac{1}{R_G}\)

or, \(R_G=\frac{I_s}{V_s}=\frac{5}{20 \times 10^{-3}}=250 \Omega\)

The option 1 is correct.