Yearly Archives: 2022

In This Post we are  providing Chapter- 10 WAVE OPTICS NCERT MOST IMPORTANT QUESTIONS for Class 12PHYSICS which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These MCQS  can be really helpful in the preparation of Board exams and will provide you with a brief knowledge of the chapter

NCERT MOST IMPORTANT QUESTIONS ON WAVE OPTICS

1.(a) State Huygens’s principle for constructing wavefronts?
(b) Using Huygens’s principle deduce the laws of reflection of light?
(c) What changes in diffraction pattern of a single slit will you observe. when the monochromatic source of light is replaced by a source of white light?
Ans.(a) According to Huygens’s principle
(1) Each source of light spreads waves in all directions.
(2) Each point on the wavefront give rise to new disturbance which produces secondary wavelets which travels with the speed of light.
(3) Only forward envelope which encloses the tangent gives the new position of wavefront.
(4) Rays are always perpendicular is the wavefront.
(b) A plane wave front AB incident at A hence every point on AB give rise to new waves. Time taken by the ray to reach from P to R







Since all the secondary wavelets takes the same time to go form the incident wavefront to the reflected wavefront so it must be independent of Q
i.e sin i– sin r = O
sin i = sin r
or i = r  law of Reflection of light
(c) (1) The diffracted light consists of different colours.
(2) It results in overlapping of different colours.

2.(a) Coloured spectrum is seen, when we look through a muslin cloth. Why?
(b) What changes in diffraction pattern of a single slit will you observe. when the monochromatic source of light is replaced by a source of white light?
Ans. (a) Muslin cloth consist of very fine threads which acts as fine slits and when light pass through it, light gets diffracted giving rise to a coloured spectrum.
(b) (i) Diffracted lights consist of different colours
(ii) It results in overlapping of different colours.

3.A slit of width ‘a’ is illuminated by light of wavelength. For what value
of ‘a’ will the :-
(i) First maximum fall at an angle of diffraction of?
(ii) First minimum fall at an angle of diffraction?
Ans.

(1) For first maximum



(2) For first minimum





 

4.(a) Derive all expression for the fringe width in young’s double slit experiment?
(b) If the two slits in young’s double slit experiment have width ratio 4:1, deduce the ratio of intensity of maxima and minima in the interference pattern?
Ans.Path difference between

——–(A)


——(1)
Using Binomial theorem expand equation. (1) and neglect higher terms

Similarity ——(2)
Substituting equation (1) & (2) in equation (A)





For bright fringes
Path difference = 




For fringe width


(b) 


Using 
 

5.Monochromatic light of wavelength 589 nm is incident from air on a water surface. What are the wavelength, frequency and speed of (a) reflected, and (b) refracted light? Refractive index o
f water is 1.33.
Ans.Wavelength of incident monochromatic light,
 = 589 nm = 
Speed of light in air, c = 3 × 108 m/s
Refractive index of water,  = 1.33
(a) The ray will reflect back in the same medium as that of incident ray. Hence, the wavelength, speed, and frequency of the reflected ray will be the same as that of the incident ray.
Frequency of light is given by the relation,


















Hence, the speed, frequency, and wavelength of the reflected light are  m/s,  Hz, and 589 nm respectively.
(b) Frequency of light does not depend on the property of the medium in which it is travelling. Hence, the frequency of the refracted ray in water will be equal to the frequency of the incident or reflected light in air.
Refracted frequency,  = Hz
Speed of light in water is related to the refractive index of water as:


Wavelength of light in water is given by the relation,




Hence, the speed, frequency, and wavelength of refracted light are  respectively.

6.In Young’s double-slit experiment using monochromatic light of wavelength , the intensity of light at a point on the screen where path difference is , is K units. What is the intensity of light at a point where path difference is  /3?
Ans.Let be the intensity of the two light waves. Their resultant intensities can be obtained as:

Where,
= Phase difference between the two waves
For monochromatic light waves,



Phase difference = 
Since path difference =  ,
Phase difference, 

Given,
I‘ = K
 …(1)
When path difference,
Phase difference, 
Hence, resultant intensity, 

Using equation (1), we can write:

Hence, the intensity of light at a point where the path difference is is units.

7. A beam of light consisting of two wavelengths, 650 nm and 520 nm, is used to obtain interference fringes in a Young’s double-slit experiment.
(a) Find the distance of the third bright fringe on the screen from the central maximum for wavelength 650 nm.
(b) What is the least distance from the central maximum where the bright fringes due to both the wavelengths coincide? 
Ans.Wavelength of the light beam, 



Wavelength of another light beam, 
Distance of the slits from the screen = D
Distance between the two slits = d
(a) Distance of the nth bright fringe on the screen from the central maximum is given by the relation,

For third bright fringe. N=3

(b) Let the nth bright fringe due to wavelength and (n – 1)th bright fringe due to wavelength coincide on the screen. We can equate the conditions for bright fringes as:

520n=650n-650
650=130n

Hence, the least distance from the central maximum can be obtained by the relation:


Note: The value of d and D are not given in the question.

8.Explain how Corpuscular theory predicts the speed of light in a medium, say, water, to be greater than the speed of light in vacuum. Is the prediction confirmed by experimental determination of the speed of light in water? If not, which alternative picture of light is consistent with experiment?
Ans.No; Wave theory
Newton’s corpuscular theory of light states that when light corpuscles strike the interface of two media from a rarer (air) to a denser (water) medium, the particles experience forces of attraction normal to the surface. Hence, the normal component of velocity increases while the component along the surface remains unchanged.
Hence, we can write the expression:
C sin i=v sin r
C sin i=v sin r
Where,
i = Angle of incidence
r = Angle of reflection
c = Velocity of light in air
v = Velocity of light in water
We have the relation for relative refractive index of water with respect to air as:

Hence, equation (i) reduces to

But, > 1
Hence, it can be inferred from equation (ii) that v > c. This is not possible since this prediction is opposite to the experimental results of c > v.
The wave picture of light is consistent with the experimental results.

9.Answer the following questions: (a) In a single slit diffraction experiment, the width of the slit is made double the original width. How does this affect the size and intensity of the central diffraction band?
(b) In what way is diffraction from each slit related to the interference pattern in a double-slit experiment?
(c) When a tiny circular obstacle is placed in the path of light from a distant source, a bright spot is seen at the centre of the shadow of the obstacle. Explain why?
(d) Two students are separated by a 7 m partition wall in a room 10 m high. If both light and sound waves can bend around obstacles, how is it that the students are unable to see each other even though they can converse easily.
(e) Ray optics is based on the assumption that light travels in a straight line. Diffraction effects (observed when light propagates through small apertures/slits or around small obstacles) disprove this assumption. Yet the ray optics assumption is so commonly used in understanding location and several other properties of images in optical instruments. What is the justification?
Ans.(a) In a single slit diffraction experiment, if the width of the slit is made double the original width, then the size of the central diffraction band reduces to half and the intensity of the central diffraction band increases up to four times.
(b) The interference pattern in a double-slit experiment is modulated by diffraction from each slit. The pattern is the result of the interference of the diffracted wave from each slit.
(c) When a tiny circular obstacle is placed in the path of light from a distant source, a bright spot is seen at the centre of the shadow of the obstacle. This is because light waves are diffracted from the edge of the circular obstacle, which interferes constructively at the centre of the shadow. This constructive interference produces a bright spot.
(d) Bending of waves by obstacles by a large angle is possible when the size of the obstacle is comparable to the wavelength of the waves.
On the one hand, the wavelength of the light waves is too small in comparison to the size of the obstacle. Thus, the diffraction angle will be very small. Hence, the students are unable to see each other. On the other hand, the size of the wall is comparable to the wavelength of the sound waves. Thus, the bending of the waves takes place at a large angle. Hence, the students are able to hear each other.
(e) The justification is that in ordinary optical instruments, the size of the aperture involved is much larger than the wavelength of the light used.
 

10.Two towers on top of two hills are 40 km apart. The line joining them passes 50 m above a hill halfway between the towers. What is the longest wavelength of radio waves, which can be sent between the towers without appreciable diffraction effects?
Ans.Distance between the towers, d = 40 km
Height of the line joining the hills, d = 50 m.
Thus, the radial spread of the radio waves should not exceed 50 km.
Since the hill is located halfway between the towers, Fresnel’s distance can be obtained as:
 = 20 km =  m
Aperture can be taken as:
a = d = 50 m
Fresnel’s distance is given by the relation,

Where,
 = Wavelength of radio waves


Therefore, the wavelength of the radio waves is 12.5 cm.Search

In This Post we are  providing Chapter- 11 DUAL NATURE OF RADIATION AND MATTER NCERT MOST IMPORTANT QUESTIONS for Class 12PHYSICS which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These MCQS  can be really helpful in the preparation of Board exams and will provide you with a brief knowledge of the chapter

NCERT MOST IMPORTANT QUESTIONS ON DUAL NATURE OF RADIATION AND MATTER

Question 1.
Write three characteristic features in photoelectric effect which cannot be explained on the basis of wave theory of light, but can be explained only using Einstein’s equation. (Delhi 2016)
Answer:
(a) (i) The maximum kinetic energy of the emitted electron should be directly proportional to the intensity of incident radiations but it is not observed experimentally. Also maximum kinetic energy of the emitted electrons should not depend upon incident frequency according to wave theory, but it is not so.
(ii) According to wave theory, threshold frequency should not exist. Light of all frequencies should emit electrons provided intensity of light is sufficient for electrons to eject.
(iii) According to wave theory, photoelectric effect should not be instantaneous. Energy of wave cannot be transferred to a particular electron but will be distributed to all the electrons present in the illuminated portion. Hence, there has to be a time lag between incidence of radiation and emission of electrons.

Question 2.
Sketch the graph showing variation of stopping potential with frequency of incident radiations for two photosensitive materials A and B having threshold frequencies v_A > v_B.
(i) In which case is the stopping potential more and why?
(ii) Does the slope of the graph depend on the nature of the material used? Explain. (All India 2016)
Answer:

Question 3.
The graphs, drawn here, are for the phenomenon of photoelectric effect.


(i) Identify which of the two characteristics (intensity/frequency) of incident light, is being kept constant in each case.

(iii) Justify the existence of a ‘threshold frequency’ for a given photosensitive surface. (Comptt. Delhi 2016)
Answer:
(i) (a) In graph 1 : intensity is being kept constant.
(b) In graph 2 : frequency is being kept constant.
(ii) (a) In graph 1 : Saturation current
(b) In graph 2 : Stopping potential.
(iii) For a given photo-sensitive surface electrons need a minimum energy to be emitted, this is called work function of the surface W.
∴ Photons energy hv should be greater/ equal to the work function.

which is justified to be called as threshold frequency.

Question 4.
Point out two distinct features observed experimentally in photoelectric effect which’ cannot be explained on the basis of wave theory of light. State how the ‘photon picture’ of light provides an explanation of these features. (Comptt. All India 2016)
Answer:
(a) (i) The maximum kinetic energy of the emitted electron should be directly proportional to the intensity of incident radiations but it is not observed experimentally. Also maximum kinetic energy of the emitted electrons should not depend upon incident frequency according to wave theory, but it is not so.
(ii) According to wave theory, threshold frequency should not exist. Light of all frequencies should emit electrons provided intensity of light is sufficient for electrons to eject.
(iii) According to wave theory, photoelectric effect should not be instantaneous. Energy of wave cannot be transferred to a particular electron but will be distributed to all the electrons present in the illuminated portion. Hence, there has to be a time lag between incidence of radiation and emission of electrons.

(b) Basic features of photon picture of electromagnetic radiation :
(i) Radiation behaves as if it is made of particles like photons. Each photon has energy E = hv and momentum p = h/λ.
(ii) Intensity of radiation can be understood in terms of number of photons falling per second on the surface. Photon energy depends only on frequency and is independent of intensity.
(iii) Photoelectric effect can be understood as the result of one to one collision between an electron and a photon.
(iv) When a photon of frequency
(v) is incident on a metal surface, a part of its energy is used in overcoming the work function and other part is used in imparting kinetic energy, so KE = h(v – v₀).

Question 5.
(i) How does one explain the emission of electrons from a photosensitive surface with the help of Einstein’s photoelectric equation?
(ii) The work function of the following metals is given : Na = 2.75 eV, K = 2.3 eV, Mo = 4.17 eV and Ni 5.15 eV. Which of these metals will not cause photoelectric emission for radiation of wavelength 3300 A from a laser source placed 1 m away from these metals? What happens if the laser source is brought nearer and placed 50 cm away? (Delhi 2017)
Answer:

Because the work function of Mo and Ni is more than the energy of the incident photons; so photoelectric emission will not take place from these two metals Mo and Ni. When the laser source is brought nearer and placed 50 cm away, the kinetic energy of photo-electrons will not change, only photoelectric current will change.

Question 6.
In the study of a photoelectric effect the graph between the stopping potential V and frequency v of the incident radiation on two different metals P and Q is shown here:
(i) Which one of the two metals has higher threshold frequency?
(ii) Determine the work function of the metal which has greater value.
(iii) Find the maximum kinetic energy of electron emitted by light of frequency 8 × 10¹⁴ Hz for this metal.
Answer:

Question 7.
Explain giving reasons for the following:
(a) Photoelectric current in a photocell increases with the increase in the intensity of the incident radiation.
(b) The stopping potential (V₀) varies linearly with the frequency (v) of the incident radiation for a given photosensitive surface with the slope remaining the same for different surfaces.
(c) Maximum kinetic energy of the photoelectrons is independent of the intensity of incident radiation.
Answer:
(a) The collision of a photon can cause emission of a photoelectron (above the threshold frequency). As the intensity increases, number of photons increases. Hence, the current increases.

hence, it depends on the frequency and not on the intensity of the incident radiation.

Question 8.
The given graph shows the variation of photocurrent for a photosensitive metal:

(a) Identify the variable X on the horizontal axis.
(b) What does the point A on the horizontal axis represent?
(c) Draw this graph for three different values of frequencies of incident radiation v₁ v₂ and v₃ (v₁ > v₂ > v₃) for same intensity.
(d) Draw this graph for three different values of intensities of incident radiation I₁ I₂ and I₃ (I₁ > I₂ > I₃) having same frequency.
Answer:
(a) ‘X’ is a collector plate potential.
(b) ‘A’ represents the stopping potential.
(c) Graph for different frequencies :

Question 9.
Draw a graph showing the variation of de Broglie wavelength of a particle of charge q and mass m with the accelerating potential. Proton and deuteron have the same de Broglie wavelengths. Explain which has more kinetic energy.
Answer:

Question 10.
(a) Draw the graph showing the variation of de Broglie wavelength of a particle of charge q and mass m with the accelerating potential.
(b) An electron and proton have the same de Broglie wavelengths. Explain, which of the two has more kinetic energy.
Answer:
(a) For Graph :


Since the mass of electron is less than that of proton, hence electron will have more kinetic energy.

In This Post we are  providing Chapter- 15 COMMUNICATION SYSTEM NCERT MOST IMPORTANT QUESTIONS for Class 12PHYSICS which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These MCQS  can be really helpful in the preparation of Board exams and will provide you with a brief knowledge of the chapter

NCERT MOST IMPORTANT QUESTIONS ON COMMUNICATION SYSTEM

Question 1.
flow will you classify’ communication systems?
Answer:
Communication systems can be classified based on the nature of source. mode of communication. type of modulation and nature of channel used.

Question 2.
What are the lips of channels used for transmission?
Answer:
i. Space communication (Broadcasting, microwave mobile etc.)
ii. Line communication (Two wire, co-axial cables, fiber optical etc.)

Question 3.
What is the length of antenna required to transmit wave of frequency 40 Hz and 40 MHz?
Answer:
The minimum length of antenna required is λ4
Velocity, c = υλ
λ = cυ

Question 4.
Which physical quantity of wave is varied in AM. FM and PM?
Answer:
In AM. the physical quantity of carrier that changes is amplitude.
In FM, the physical quantity of carriers that change is frequency.
In PM, the physical quantity of carrier that changes is a phase.

Question 5.
What are the advantages and limitations of AM and FM?
Answer:

Question 6.
What is ground wave propagation?
Answer:
Ground wave follows curvature of the earth and has carrier frequencies up to 2MHz. e.g. AM radio.
Ground waves progress along the surface of the earth and must be vertically polarized to prevent short circuiting the electric equipments. A wave induces currents in the ground over which it passes and thus loses some energy by absorption. This is made up by energy diffracted downwards from the upper portion of the wavefront.

There is another way also by which the ground waves get attenuated. Because of diffraction, the wavefront gradually tilts over, as shown in the figure. As the wave propagates over the earth, the tilt increases and this tilt causes greater short-circuiting of the electric component of the wave. Hence there is a reduction in the field strength. Eventually, at some distance from the antenna, the wave gets weakened and dies off. The maximum range of such a transmitter depends on its frequency and power. The ground wave propagation is effective only at VLF.

Question 7.
Through which atmospheric layer. does the propagation take place in ground. space and sky communications?
Answer:
Ground wave – Troposphere
Space wave – Troposphere
Sky wave – Ionosphere

Question 8.
For establishing a communication between a transmitting and receiving station, a physical medium is used.
Answer:
(a) Name the two principal classes of communication based on the physical medium used for propagation.
(b) Construct a table showing advantages and one practical application each for the two wire, coaxial cable and optic fiber communication.
(c) in cable TV transmission usually channel in UHF band carries relatively more noise, compared to VHF band. Justify
Answer:
(a) Line communication and space communication.

(c) At higher frequency, radiation loss is high.

Question 9.
Schematic diagram for three types of satellite orbits are shown below and named as A.B.C.
Answer:

(a) Identify the polar orbit and give its approximate height from earth.
(b) Give the criteria for selecting frequency of em wave to be used in photographs from satellites.
(c) A satellite T V company attempts to use 25,000 kHz for up linking signal to a sat ellite. Say whether they have selected apt frequency. Justify.
Answer:
(a) Orbit C. Its height is about 1000 km.
(b) i. Nature of the atmosphere.
ii. Reluctance of the object.
(c) No. Because frequency below 20 MHz will undergo total internal reflection at the ionosphere.

Question 10.
The following diagrams represent some of the modulated signals.



Which among is following correct
a. i only
b. ii only
c. iii only
d. both i and ii
Answer:
d. both i and ii

In This Post we are  providing Chapter- 14 SEMICONDUCTOR ELECTRONIC : MATERIAL, DEVICES AND SIMPLE CIRCUITS NCERT MOST IMPORTANT QUESTIONS for Class 12PHYSICS which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These MCQS  can be really helpful in the preparation of Board exams and will provide you with a brief knowledge of the chapter

NCERT MOST IMPORTANT QUESTIONS ON SEMICONDUCTOR ELECTRONIC : MATERIAL, DEVICES AND SIMPLE CIRCUITS

1.Draw a pn junction with reverse bias? Which biasing will make the resistance of a
p-n-junction high?
Ans.

Reverse biasing will make the resistance high as it will not allow the current to pass.

2.Write the truth table for the following combination of gates?

Ans.

3.Draw the voltage current characteristics of a zener diode?
Ans.

4.For a extrinsic semiconductor, indicate on the energy band diagram the donor and
acceptor levels?
Ans. N-type Extrinsic Semiconductor P-type Extrinsic Semiconductor

5.What do you mean by depletion region and potential barrier in junction diode?
Ans.A layer around the junction between p and n-sections of a junction diode where charge carriers electrons and holes are less in number is called depletion region. The potential difference created across the junction due to the diffusion of charge carriers across the junction is called potential barrier.

6.A transistor has a current gain of 30. If the collector resistance is 6k, input resistance is 1k, calculate its voltage gain?
Ans.Given 


Voltage gain = current gain Rgain
Voltage gain = 306 = 180

7. What are the advantages and disadvantages of semiconductor devices over vacuum tubes?
Ans.Advantages – Semiconductor devices are very small in size as compared to the vacuum tubes. It requires low voltage for their operation
Disadvantage – Due to the rise in temperature and by applying high voltage it can be damaged.

8.The base of a transistor is lightly doped. Explain why?
Ans.In a transistor, the majority carries form emitter region moves towards the collector region through base. If base is made thick and highly doped, majority carriers will combine with the other carriers within the base and only few is collected by the collector which leads to small output collector current. Thus in order to have large output collector current, base is made thin and lightly doped.
 

9. Determine the currents through resistance R of the circuits (i) and (ii) when similar diodes are connected as shown in the figure.

Ans.In figure (i) are forward biased

In figure (ii) is forward biased but is reverse biased due to which  offers infinite resistance

10.What do you mean by hole in a circuit? Write its two characteristics?
Ans.A vacancy created in a covalent bond in a semiconductor due to the release of electron is known as hole in a semiconductor.
Characteristics of hole
(i) Hole is equivalent to a positive electronic charge.
(ii) Mobility of hole is less than that of an electron

In This Post we are  providing Chapter- 12 ATOMS NCERT MOST IMPORTANT QUESTIONS for Class 12PHYSICS which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These MCQS  can be really helpful in the preparation of Board exams and will provide you with a brief knowledge of the chapter

NCERT MOST IMPORTANT QUESTIONS ON ATOMS

Question 1.
A hydrogen atom in its excited state emits radiations of wavelengths 1218 Å and 974.3 Å when it finally comes to the ground state. Identify the energy levels from where transitions occur. Given Rydberg constant R = 1.1 × 10⁷ m^-1. Also, specify the spectral series to which these lines belong.
Answer:
We know that


On solving n = 4
Lyman series.

Question 2.
Obtain the first Bohr’s radius and the ground state energy of a muonic hydrogen atom, i.e. an atom where the electron is replaced by a negatively charged muon (μ^–) of mass about 207 me that orbits around a proton. (Given for hydrogen atom, the radius of first orbit and ground state energy are 0.53 × 10^-10m and – 13.6 eV respectively)
Answer:
In Bohr’s model of hydrogen atom the radius of n^th orbit is given by
r_n = n2h24π2e2mek

As n = 1
Therefore, we have 1

Question 3.
The electron in a given Bohr orbit has a total energy of – 1.5 eV. Calculate Its
(a) kinetic energy.
(b) potential energy.
(C) the wavelength of radiation emitted, when this electron makes a transition to the ground state.
(Given Energy in the ground state = – 13.6 eV and Rydberg’s constant = 1.09 × 1o⁷ m^-1)
Answer:
Total energy of the electron In a Bohr’s orbit is – 1.5 eV

We know that kinetic energy of the electron in any orbt is half of the potential energy in magnitude and potential energy is negative
(a) Total energy = kinetic energy + potential energy
– 1.5 = E_k – 2E_k
1.5 = E_k

(b) E_p = – 2 × 1.5 = – 3 eV

(c) Energy released when the transition of this electron takes place from this orbit to the ground state
= – 1.5 – (- 13.6)
= 12.1 eV
= 12.1 × 1.6 × 10¹⁹ = 1.936 × 10^-18 J

Let be the wavelength of the Light emitted then,

Question 4.
In a Geiger—Marsden experiment, calculate the distance of closest approach to the nucleus of Z = 80, when an a-particle of 8 MeV energy Impinges on it before It comes momentarily to rest and reverses its direction. How will the distance of the closest approach be affected when the kinetic energy of the a-particle is doubled?
Answer:
Given Z = 80, E = 8 MeV = 8 × 10⁶ × 1.6 × 10¹⁹ J, r_o =?
Using the expression

When the kinetic energy of a-particle has doubled the distance of the closest approach becomes half its previous value, i.e. 1.44 × 10¹⁴ m

Question 5.
The ground state energy of the hydrogen atom is – 13.6 eV. If an electron makes a transition from an energy level – 0.85 eV to – 3.4 eV, calculate the wavelength of the spectral line emitted. To which series of hydrogen spectrum does this wavelength belong?
Answer:
Energy released = – 0.85 – (- 3.4) = 2.55 eV = 2.55 × 1.6 × 10^-19 J
Using E = hc/λ we have
λ = hcE=6.62×10−34×3×1082.55×1.6×10−19 = 4.87 × 10^-7 m

It belongs to the Balmer series.

Question 6.
A hydrogen atom in the third excited state de-excites to the first excited state. Obtain the expressions for the frequency of radiation emitted in this process.
Also, determine the ratio of the wavelengths of the emitted radiations when the atom de-excites from the third excited state to the second excited state and from the third excited state to the first excited state.
Answer:
We know that


Question 7.
Obtain the expression for the ratio of the de Broglie wavelengths associated with the electron orbiting in the second and third excited states of the hydrogen atom.
Answer:
We know that
2πr = nλ ….(i)
For the second excited state (n = 3)
r = 0.529(n)² A = 0.529(3)²

Putting in (i) we get 2π(0.529)(3)² = 3 λ₂

For third excited state n = 4
r = 0.529 (4)²
Putting in (i) we get 2π (0.529)(4)² = 4 λ₃

Question 8.
(a) The radius of the innermost electron orbit of a hydrogen atom is 5.3 × 10^-10 m. Calculate its radius in n = 3 orbit.
(b) The total energy of an electron in the first excited state of the hydrogen atom is 3.4 eV. Find out its (i) kinetic energy and (ii) potential energy in this state.
Answer:
(a) The radius is given by
r_n = n2h24π2me2k = n²ao

where n is the number of orbit, hence r3 = 5.3 × 10^-10 × 32 = 4.77 × 10^-9 m

(b) Total kinetic energy = + 3.4 eV
Total potential energy = – 6.8 eV

Question 9.
Given the ground state energy E0 = – 13.6 eV and Bohr radius a0 = 0.53 Å. Find out how the de Broglie wavelength associated with the electron orbiting in the ground state would change when it jumps into the first excited state.
Answer:
The de-Broglie wavelength is given by 2πrn = nλ.
In ground state, n = 1 and r_o = 0.53 Å, therefore, λ_o = 2 × 3.14 × 0.53 = 3.33 Å

In first excited state, n = 2 and r₁ = 4 × 0.53 Å = 2.12 Å, therefore,
r₁ = (2 × 3.14 × 2.12)/2 = 6.66Å

Therefore, λ₁ – λ_o = 6.66 – 3.33 = 3.33 Å.
In other words, the de-Broglie wavelength becomes double.

Question 10.
When is the H_α line in the emission spectrum of hydrogen atom obtained? Calculate the frequency of the photon emitted during this transition.
Answer:
Hα is obtained when n_i = 3 and n_f = 2

In This Post we are  providing Chapter- 13 NUCLEI NCERT MOST IMPORTANT QUESTIONS for Class 12PHYSICS which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These MCQS  can be really helpful in the preparation of Board exams and will provide you with a brief knowledge of the chapter

NCERT MOST IMPORTANT QUESTIONS ON NUCLEI

Question 1.
Calculate the binding energy per nucleon of Fe⁵⁶₂₆ Given m_Fe = 55.934939 u, m_n = 1.008665 u and m_p = 1.007825 u
Answer:
Number of protons Z = 26
Number of neutrons (A – Z) = 30
Now mass defect is given by
Δm = Z m_p + (A – Z)m_n – M
Δm = 26 × 1.007825 + 30 × 1.008665 – 55.934939
= 0.528461 u

Therefore binding energy
BE = Δm × 931 MeV = 0.528461 × 931
= 491.99 MeV

BE/nucleon = 491.99/56 = 8.785 MeV

Question 2.
The activity of a radioactive element drops to one-sixteenth of its initial value in 32 years. Find the mean life of the sample.
Answer:

Or
32/T = 4 or 7 = 32 / 4 = 8 years.
Therefore mean life of the sample is τ = 1.44 7 = 1.44 × 8 = 11.52 years.

Question 3.
A radioactive sample contains 2.2 mg of pure 116C which has a half-life period of 1224 seconds. Calculate (i) the number of atoms present initially and (ii) the activity when 5 pg of the sample will be left.
Answer:
Mass of sample = 2.2 pg
Now 11 g of the sample contains 6.023 × 10²³ nuclei, therefore the number of nuclei in 2.2 mg = 2.2 × 10^-3 g are

Question 4.
The half-life of 238 92U is 4.5 × 10⁹ years. Calculate the activity of 1 g sample of ₉₂²³⁸U.
Answer:
Given T = 4.5 × 10⁹ years.
Number of nuclei of U in 1 g
= N = 6.023×1023238 = 2.5 × 10²¹

Therefore activity

Question 5.
The decay constant for a given radioactive sample is 0.3456 per day. What percentage of this sample will get decayed in a period of 4 days?
Answer:
Given λ = 0.3456 day^-1
or
T_1/2 = 0.693/λ = 0. 693/ 0.3456 = 2.894 days, t = 4 days.

Let N be the mass left behind, then N = N_oe^-λt
or
N = N_o e^{-0 3456 × 4}
or
N = N₀ e^{-1 3824} = N_o × 0.25

Therefore the percentage of undecayed is

Question 6.
It is observed that only 6.25 % of a given radioactive sample is left undecayed after a period of 16 days. What is the decay constant of this sample per day?
Answer:
Given N/N_o = 6.25 %, t = 16 days, λ = ?

Or
16/ T = 4 or T = 4 days.

Therefore λ = 1/T = 1/4 = 0.25 day^-1

Question 7.
A radioactive substance decays to 1/32^th of its initial value in 25 days. Calculate its half-life.
Answer:
Given t = 25 days, N = N_o / 32, using

Or
25/7= 5 or T= 25 / 5 = 5 days.

Question 8.
The half-life of a radioactive sample is 30 s.
Calculate
(i) the decay constant, and
Answer:
Given T₁/2 = 30 s, N = 3N_o / 4, λ = ?, t = ?
(i) Decay constant
λ = 0.693T1/2=0.69330 = 0.0231 s^-1

(ii) time taken for the sample to decay to 3/4 th of its initial value.
Answer:
Using N = N_oe^-λt we have

Question 9.
The half-life of 14 6C is 5700 years. What does it mean?
Two radioactive nuclei X and Y initially contain an equal number of atoms. Their half-lives are 1 hour and 2 hours respectively. Calculate the ratio of their rates of disintegration after 2 hours.
Answer:
It means that in 5700 years the number of nuclei of carbon decay to half their original value.
Given N_ox = N_oY, T_X = 1 h, T_Y = 2 h, therefore
λXλY=21 = 2

Now after 2 hours X will reduce to one- fourth and Y will reduce to half their original value.
If activities at t = 2 h are R_x and R_y respectively, then

Thus their rate of disintegration after 2 hours is the same.

Question 10.
A star converts all its hydrogen to helium achieving 100% helium composition. It then converts helium to carbon via the reaction.

The mass of the star is 5 × 10³² kg and it generates energy at the rate of 5 × 10³⁰ watt. How long will it take to convert all the helium to carbon at this rate?

Answer:

As 4 × 10^-3 kg of He consists of 6.023 × 10²³ He nuclei so 5 × 10³² kg He will contain
6.023×1023×5×10324×10−3 = 7.5 × 10⁵⁸ nuclei

Now three nuclei of helium produce 7.27 × 1.6 × 10^-13 J of energy
So all nuclei in the star will produce
E = 7.27×1.6×10−133 × 7.5 × 10⁵⁸
= 2.9 × 10⁴⁶ J

As power generated is P = 5 × 10³⁰ W, therefore time taken to convert all He nuclei into carbon is
t = EP=2.9×10465×1030 = 5.84 × 10¹⁵ s
or
1.85 × 10⁸ years

In This Post we are  providing Chapter-4 MOVING CHARGES AND MAGNETISM NCERT MOST IMPORTANT QUESTIONS for Class 12PHYSICS which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These MCQS  can be really helpful in the preparation of Board exams and will provide you with a brief knowledge of the chapter

NCERT MOST IMPORTANT QUESTIONS ON MOVING CHARGES AND MAGNETISM

Question 1.
Draw a schematic diagram of a cyclotron. Explain its underlying principle and working, stating clearly the function of the electric and magnetic field applied on a charged particle. Deduce an expression for the period of revolution and show that it does not depend on the speed of the charged particle.
Answer:
Principle : When a positively charged particle is made to move again and again in a high frequency electric field, it gets accelerated and acquires sufficiently large amount of energy.

Working : Suppose a positive ion, say a proton, enters the gap between the two dees and finds dee D₁ to be negative. It gets accelerated towards dee D₁. As it enters the dee D₁, it does not experience any electric field due to shielding
effect of the metallic dee. The perpendicular magnetic field throws it into a circular path.

At the instant the proton comes out of dee D₁. It finds dee D₁ positive and dee D₂ negative. It now gets accelerated towards dee D₂. It moves faster through dee D₂ describing a larger semicircle than before. Thus if the frequency of the applied voltage is kept exactly the same as the frequency of the revolution of the proton, then everytime the proton reaches the gap between the two dees, the electric field is reversed and proton receives a push and finally it acquires very high energy. This proton follows a spiral path. The accelerated proton is ejected through a window by a deflecting voltage and hits the target.
Centripetal force is provided by magnetic field to charged particle to move in a circular back.

Question 2.
Draw a schematic sketch of a cyclotron. Explain briefly how it works and how it is used to accelerate the charged particles.
(i) Show that time period of ions in a cyclotron is independent of both the speed and radius of circular path.
(ii) What is resonance condition? How is it used to accelerate the charged particles? (All India 2017)
Answer:
(i) Principle : When a positively charged particle is made to move again and again in a high frequency electric field, it gets accelerated and acquires sufficiently large amount of energy.

Working : Suppose a positive ion, say a proton, enters the gap between the two dees and finds dee D₁ to be negative. It gets accelerated towards dee D₁. As it enters the dee D₁, it does not experience any electric field due to shielding effect of the metallic dee. The perpendicular magnetic field throws it into a circular path.

At the instant the proton comes out of dee D₁. It finds dee D₁ positive and dee D₂ negative. It now gets accelerated towards dee D₂. It moves faster through dee D₂ describing a larger semicircle than before. Thus if the frequency of the applied voltage is kept exactly the same as the frequency of the revolution of the proton, then everytime the proton reaches the gap between the two dees, the electric field is reversed and proton receives a push and finally it acquires very high energy. This proton follows a spiral path. The accelerated proton is ejected through a window by a deflecting voltage and hits the target.
Centripetal force is provided by magnetic field to charged particle to move in a circular back.

(ii) The frequency v_a of the applied voltage is adjusted so that the polarity of the dees is reversed in the same time that it takes the ions to complete one-half of the revolution. The requirement v_a = v_c is called the resonance condition.
The phase of the supply is adjusted so that when the positive ions arrive at the edge of D₁, D₂ is at a lower potential and the ions are accelerated across the gap.

Question 3.
(a) Two straight long parallel conductors carry currents I₁ and I₂ in the same direction. Deduce the expression for the force per unit length between them.
Depict the pattern of magnetic field lines around them.
(b) A rectangular current carrying loop EFGH is kept in a uniform magnetic field as shown in the figure.
(i) What is the direction of the magnetic moment of the current loop?
(ii) When is the torque acting on the loop
(A) maximum,
(B) zero?

Answer:
Consider two infinitely long parallel conductors carrying current I₁ and I₂ in the same direction.
Let d be the distance of separation between these two conductors.

Hence, force is attractive in nature.
Ampere : Ampere is that current which is if maintained in two infinitely long parallel conductors of negligible cross-sectional area separated by 1 metre in vacuum causes a force of 2 × 10^-7 N on each metre of the other wire.
Then current flowing is 1A


(i) Magnetic moment will be out of the plane from the surface HEFG.
(ii) Torque
(A) Torque is maximum when MII B i.e., when it gets rotated by 90°.
(B) Torque is minimum when M and B are at 270° to each other.

Question 4.
(a) With the help of a diagram, explain the principle and working of a moving coil galvanometer.
(b) What is the importance of a radial magnetic field and how is it produced?
(c) Why is it that while using a moving coil galvanometer as a voltmeter a high resistance in series is required whereas in an ammeter a shunt is used? (All India)
Answer:
(a) Principle : “If a current carrying coil is freely suspended/pivoted in a uniform magnetic field, it experiences a deflecting torque.”
Working: As the pivoted coil is placed in a radial magnetic field, hence on passing current I through it, a deflecting torque acts on the coil which is given by, τ = NAIB

The spring S_p attached to the coil provides the counter torque and in equilibrium state balances the deflecting torque. If φ is steady angular deflection then counter torque is kφ.
…where [k = torsional constant of the spring

In equilibrium state,

Thus, deflection is directly proportional to the current flowing in the coil.
(a) (i) Uniform radial magnetic field. It keeps the magnetic field line normal to the area vector of the coil.
(ii) Soft iron core in galvanometer. The cylindrical soft iron core, when placed inside the coil of a galvanometer, makes the magnetic field stronger and radial in the space between it and pole pieces, such that whatever the position of the rotation of the coil may be, the magnetic field is always parallel to its plane.

(b) (i) Current sensitivity is defined as the deflection produced in the galvanometer when unit current is passed through its coil.

(ii) Voltage sensitivity is defined as the deflection produced in the galvanometer when unit voltage is applied across the coil of the galvanometer.

…where [R = Resistance of the coil

does not necessarily increase the voltage sensitivity. It may be affected by the resistance used.

(b) For radial magnetic field, sin θ = 1,
so torque τ = NIAB.
Thus when radial magnetic field is used, the deflection of the coil is proportional to the current flowing through it. Hence a linear scale can be used to determine the deflection of the coil.

(c) A high resistance is joined in series with a galvanometer so that when the arrangement (voltmeter) is used in parallel with the selected section of the circuit, it should draw least amount of current. In case voltmeter draws appreciable amount of current, it will disturb the original value of potential difference by a good amount.

To convert a galvanometer into ammeter, a shunt is used in parallel with it so that when the arrangement is joined in series, the maximum current flows through the shunt, and thus the galvanometer is saved from its damage, when the current is passed through ammeter.

Question 5.
(a) Derive an expression for the force between two long parallel current carrying conductors.

(b) Use this expression to define S.I. unit of current.
(c) A long straight wire AB carries a current I. A proton P travels with a speed v, parallel to the wire, at a distance d from it in a direction opposite to the current as shown in the figure. What is the force experienced by the proton and what is its direction?
Answer:
(a) For (a) and (b) :
Consider two infinitely long parallel conductors carrying current I₁ and I₂ in the same direction.
Let d be the distance of separation between these two conductors.

Hence, force is attractive in nature.
Ampere : Ampere is that current which is if maintained in two infinitely long parallel conductors of negligible cross-sectional area separated by 1 metre in vacuum causes a force of 2 × 10^-7 N on each metre of the other wire.
Then current flowing is 1A


(c) Force experienced by the proton,

As magnetic field due to the current carrying wire is directed into the plane of the paper (θ = 90°)

Force is directed away from the current carrying wire or in the right direction of observer.

Question 6.
State Biot-Savart law, giving the mathematical expression for it.
Use this law to derive the expression for the magnetic field due to a circular coil carrying current at a point along its axis.
How does a circular loop carrying current behave as a magnet? (Delhi 2011)
Answer:
According to Biot-Savart’s law, “magnetic field acting at a particular point due to current carrying element is proportional to the division of cross product of current element and position vector of point where the field is to be calculated from the current element to the cube of the distance between current element and the point where the field is to be calculated”.


Magnetic field on the axis of circular current loop :

As in a special case we may obtain the field at the centre of the loop. Here x = 0, and we obtain

In a current loop, both the opposite faces behave as opposite poles, making it a magnetic dipole. One side of the current carrying coil behaves like the N-pole and the other side as the S-pole of a magnet.

Question 7.
With the help of a labelled diagram, state the underlying principle of a cyclotron. Explain clearly how it works to accelerate the charged particles. Show that cyclotron frequency is independent of energy of the particle. Is there an upper limit on the energy acquired by the particle? Give reason.

Answer:
Principle : When a positively charged particle is made to move again and again in a high frequency electric field, it gets accelerated and acquires sufficiently large amount of energy.

Working : Suppose a positive ion, say a proton, enters the gap between the two dees and finds dee D₁ to be negative. It gets accelerated towards dee D₁. As it enters the dee D₁, it does not experience any electric field due to shielding effect of the metallic dee. The perpendicular magnetic field throws it into a circular path.

At the instant the proton comes out of dee D₁. It finds dee D₁ positive and dee D₂ negative. It now gets accelerated towards dee D₂. It moves faster through dee D₂ describing a larger semicircle than before. Thus if the frequency of the applied voltage is kept exactly the same as the frequency of the revolution of the proton, then everytime the proton reaches the gap between the two dees, the electric field is reversed and proton receives a push and finally it acquires very high energy. This proton follows a spiral path. The accelerated proton is ejected through a window by a deflecting voltage and hits the target.
Centripetal force is provided by magnetic field to charged particle to move in a circular back.

Yes, there is an upper limit. The increase in the kinetic energy of particles is qv. Therefore, the radius of their path goes on increasing each time, their kinetic energy increases. The lines are repeatedly accelerated across the dees, untill they have the required energy to have a radius approximately that of the dees. Hence, this is the upper limit on the energy required by the particles due to definite size of dees.

Question 8.
(a) State the principle of the working of a moving coil galvanometer, giving its labelled diagram.
(b) “Increasing the current sensitivity of a galvanometer may not necessarily increase its voltage sensitivity.” Justify this statement
(c) Outline the necessary steps to convert a galvanometer of resistance RG into an ammeter of a given range. (All India 2011)
Answer:
(a)
Principle : “If a current carrying coil is freely suspended/pivoted in a uniform magnetic field, it experiences a deflecting torque.”
Working: As the pivoted coil is placed in a radial magnetic field, hence on passing current I through it, a deflecting torque acts on the coil which is given by, τ = NAIB

The spring S_p attached to the coil provides the counter torque and in equilibrium state balances the deflecting torque. If φ is steady angular deflection then counter torque is kφ.
…where [k = torsional constant of the spring

In equilibrium state,

Thus, deflection is directly proportional to the current flowing in the coil.
(a) (i) Uniform radial magnetic field. It keeps the magnetic field line normal to the area vector of the coil.
(ii) Soft iron core in galvanometer. The cylindrical soft iron core, when placed inside the coil of a galvanometer, makes the magnetic field stronger and radial in the space between it and pole pieces, such that whatever the position of the rotation of the coil may be, the magnetic field is always parallel to its plane.
(b) (i) Current sensitivity is defined as the deflection produced in the galvanometer when unit current is passed through its coil.

(ii) Voltage sensitivity is defined as the deflection produced in the galvanometer when unit voltage is applied across the coil of the galvanometer.

…where [R = Resistance of the coil

does not necessarily increase the voltage sensitivity. It may be affected by the resistance used.

(b) Since vs=IsR increase in current sensitivity may not necessarily increase the voltage sensitivity. It may be affected by the resistance used.

(c) Conversion of galvanometer into ammeter: By just connecting a low resistance known as shunt in parallel to the galvanometer, it can be converted into an ammeter.

Let G = resistance of the galvanometer.
I_g = the current with which galvanometer gives full scale deflection.
S = shunt resistance
I – I_g = current through the shunt.
As the galvanometer and shunt are connected in parallel,
Potential difference across the galvanometer = Potential difference across the shunt

Question 9.
(a) Write the expression for the force, F→, acting on a charged particle of charge ‘q’, moving with a velocity latex]\overrightarrow{\mathbf{v}}[/latex] in the presence of both electric field E→ and magnetic field B→. Obtain the condition under which the particle moves undeflected through the fields.
(b) A rectangular loop of size l × b carrying a steady current I is placed in a uniform magnetic field B→. Prove that the torque τ⃗  acting on the loop is given by τ⃗ =m⃗ ×B→, where m→ is the magnetic moment of the loop. (All India 2011)
Answer:
(a) A charge q in an electric field E→ experiences the electric force, F→e=qE→

This force acts in the direction of field E→ and is independent of the velocity of the charge.

The magnetic force experienced by the charge q moving with velocity v→ in the magnetic field B is given by

This force acts perpendicular to the plane of V→ and B→ and depends on the velocity v→ of the charge.

The total force, or the Lorentz force, experienced by the charge q due to both electric and magnetic field is given by

Hence, A stationary charged particle does not experience any force in a magnetic field. (b) Torque on a current loop in a uniform magnetic field.
Let I = Current flowing through the coil PQRS

Its magnitude is, F3 = IaB sin(90° + 0)

Question 10.
(a) Explain, giving reasons, the basic difference in converting a galvanometer into
(i) a voltmeter and
(ii) an ammeter.
(b) Two long straight parallel conductors carrying steady currents I₁ and I₂ are separated by a distance’d’ Explain briefly, with the help of a suitable diagram, how the magnetic field due to one conductor acts on the other. Hence deduce the expression for the force acting between the two conductors. Mention the nature of this force.
Answer:
(a) (i) Voltmeter is connected in parallel with the circuit element across which the potential difference is intended to be measured.

A galvanometer can be converted into a voltmeter by connecting a higher resistance in series with it. The value of this resistance is so adjusted that only current I which produces full scale deflection in the galvanometer, passes through the galvanometer.

(ii) A galvanometer can be converted into an ammeter by connecting a low value
resistance in parallel with it.


(b)
Consider two infinitely long parallel conductors carrying current I₁ and I₂ in the same direction.
Let d be the distance of separation between these two conductors.

Hence, force is attractive in nature.
Ampere : Ampere is that current which is if maintained in two infinitely long parallel conductors of negligible cross-sectional area separated by 1 metre in vacuum causes a force of 2 × 10^-7 N on each metre of the other wire.
Then current flowing is 1A

In This Post we are  providing Chapter- 7 ALTERNATING CURRENT NCERT MOST IMPORTANT QUESTIONS for Class 12 PHYSICS which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These MCQS  can be really helpful in the preparation of Board exams and will provide you with a brief knowledge of the chapter

NCERT MOST IMPORTANT QUESTIONS ON ALTERNATING CURRENT

Question 1.
The figure shows how the reactance of an Inductor varies with frequency. (a) Calculate the value of the Inductance of the Inductor using Information given In the graph. (b) If this Inductor is connected In senes to a resistor of 8 ohms, find what would be the impedance at 300 Hz.

Answer:
(a) We know that X_L = 2πfL or L = XL2πf.
Now slope of the graph is
XLf=8−6400−300=2100 = 0.02

Therefore L is L = XL2πf=0.022×3.14 = 0.0032 H

(b) NowR = 8 ohm, f = 300 Hz, Z = ?
Now Z = R2+X2L−−−−−−−√ . Therefore we have
Z = R2+X2L−−−−−−−√ = (8)2+(6)2−−−−−−−−−√ = 10 Ω

Question 2.
A 25.0 μF capacitor, a 0.10-henry Inductor, and a 25.0-ohm resistor are connected in series with an ac source whose emf is E= 310 s. In 314 t (i) what is the frequency of the .mf? (ii) Calculate (a) the reactance of the circuit (b) the Impedance of the circuit and (C) the current in the circuit.
Answer:
Given C = 25.0 μF, L = 0.10 henry, R = 25.0 ohm, E_o = 310V,
Comparing with the equation E = E_o sin ωt,
we have
(i) ω = 314 or f = 50 Hz

Question 3.
A sinusoidal voltage V = 200 sin 314 t Is applied to a resistor of 10 ohms. Calculate (i) rms value of current (ii) rms value of voltage and (iii) power dissipated as heat in watt.
Answer:
V_o = 200 V, ω = 314 rads^-1 , V_rms = ?, l_rms = ?, P = ?

Question 4.
Find the inductance of the inductor used in series with a bulb of resistance 10 ohms connected to an ac source of 80 V, 50 Hz. The power factor of the circuit is 0.5. Also, calculate the power dissipation in the circuit.
Answer:
Given R = 10 ohm, V = 80 Hz, f = 50 Hz, cos Φ = 0.5 , L = ?, P = ?
Using the formula
cos Φ = RR2+X2L√
or
0.25(R² + X²) = R²
Or
O.25X_L² = O.75R²
or
X_L = 17.32 ohm

Now using X_L = 2πfL we have
L = XL2πf=17.322×3.14×50 = 0.055 H

Now Z = R2+X2L−−−−−−−√
= (10)2+(17.32)2−−−−−−−−−−−−−√ = 20 Ω

Now P_av = l_rms V_rms cos Φ = V2mZ × cos Φ
or
Pav = (80)22×20 × 0.5 = 160 W

Question 5.
When an alternating voltage of 220 V is applied across a device X, a current of 0.5 A flows through the circuit and Is in phase with the applied voltage. When the same voltage is applied across a device Y, the same current again flows through it, but it leads the voltage by π/2. If element ‘X’ is a pure resistor of 100 ohms,
(a) name the circuit element ‘Y’ and
Answer:
The element Y is a capacitor.

(b) calculate the rms value of current, if rms value of voltage is 141 V.
Answer:
The value of Xc is obtained as below

X_C = VI=2200.5 = 440 ohm

Therefore impedance of the circuit
Z = R2+X2C−−−−−−−√=(100)2+(440)2−−−−−−−−−−−−√ = 451.2 ohm

Therefore rms value of current V 141
l = VrmsZ=141451.2 = 0.3125 A

Question 6.
When an alternating voltage of 220 V is applied across a device X, a current of 0. 5 A flows through the circuit and is in phase with the applied voltage. When the same voltage is applied across a device V, the same current again flows through it, but it lags the voltage by π/2.
(a) Name the devices X and Y.
Answer:
The element X is a resistor and Y is an inductor.

(b) Calculate the current flowing through the circuit when the same voltage is applied across the series combination of the two devices X and Y.
Answer:
Now both R and XL are the same and are given by
R = X_L = 2200.5 = 440 ohm

Hence impedance of the circuit
Z = R2+X2L−−−−−−−√
= (440)2+(440)2−−−−−−−−−−−−√
= 622.2 ohm

Therefore current flowing through the circuit is
l = VZ = 220622.2 = 0.353 A

Question 7.
A 15.0 μF capacitor Is connected to a 220 V, 50 Hz source. Find the capacitive reactance and the current (rms and peak) In the circuit. If the frequency Is doubled, what happens to the capacitive reactance and the current?
Answer:
Given C= 15.0 μF= 15 × 10^-6 F, V= 220 V,
f = 50 HZ, X_C = ? l_m = ?

The capacitive reactance is

Now l_rms = VrmsXc=220212 = 1.04 A

Peak value of current
l_m = 2–√ × lrms = 1.4.1 × 1.04 = 1.47 A

This current oscillates between + 1.47A and – 1.47 A, and is ahead of the voltage by π/2.

If the frequency is doubled, the capacitive reactance is halved and consequently, the current is doubled.

Question 8.
A sinusoidal voltage of peak value 283 V and frequency 50 Hz is applied to a series LCR circuit in which R = 3 Ω, L = 25.48 mH and C = 796 μF. Find
(a) the impedance of the circuit;
(b) the phase difference between the voltage across the source and the current;
(c) the power dissipated in the circuit; and
(d) the power factor.
Answer:
Given V_m = 283 V, f = 50 Hz, R = 3 Ω
L = 25.48 mH, and C = 796 μF.
(a) To find the impedance of the circuit, we first calculate XL and Xc
X_L = 2πfL = 2 × 3.14 × 50 × 25.48 × 10^-3 = 8 Ω
X_C = 12πfC
= 12×3.14×50×796×10−6 = 4 Ω

Therefore impedance of the circuit is
Z = R2+(XL−XC)2−−−−−−−−−−−−−−√
= 32+(8−4)2−−−−−−−−−−√
= 5 Ω

(b) Phase difference
Φ = tan^-1XL−XcR = tan-1 8−43 = 53.1°
Since Φ is positive therefore voltage leads current by the above phase.

(c) The power dissipated in the circuit is
P = V2rmsZ=V2m2Z=(283)22×5 = 8008.9 W
(d) power factor = cos Φ = cos 53.1° = 0.6

Question 9.
A capacitor and a resistor are connected in series with an ac source. If the potential difference across the C, R is 120 V and 90 V respectively, and if rms current of the circuit is 3 A, calculate the (i) impedance and (ii) power factor of the circuit.
Answer:
Given V_C = 120 V, V_R = 90 V, f = 3 A. and R = 90/3 = 30 ohm ,

Effective voltage in the circuit
V = V2C+V2R−−−−−−−√=(120)2+(90)2−−−−−−−−−−−√= 150 V .
(i) Therefore impedance of the circuit
Z = Vl=1503 = 50 Ω.

(ii) Now power factor of the circuit is
cos Φ = RZ=3050 = 0.6

Question 10.
An inductor 200 mH, a capacitor C, and a resistor 10 ohm are connected in series with 100 V, 50 Hz ac source. If the current and the voltage are in phase with each other, calculate the capacitance of the capacitor.
Answer:
When current and voltage are in phase then X_L = X_C

In This Post we are  providing Chapter-9 RAY OPTICS AND OPTICAL INSTRUMENTS NCERT MOST IMPORTANT QUESTIONS for Class 12 PHYSICS which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These MCQS  can be really helpful in the preparation of Board exams and will provide you with a brief knowledge of the chapter

NCERT MOST IMPORTANT QUESTIONS ON RAY OPTICS AND OPTICAL INSTRUMENT

Question 1.
A convex lens made up of a glass of refractive index 1.5 is dipped, In turn, In
(a) a medium of refractive index 1.65,
Answer:
When dipped in the medium of refractive index 1.65, it will behave as a concave lens and when dipped in the medium of refractive index 1.33, it will behave as a convex lens.

(b) a medium of refractive index 1.33.
(i) Will it behave as a converging or a diverging lens in the two cases?
Answer:
Its focal length in another medium is given by

Thus f_m = -5.5 f_a, i.e. focal length increases and becomes negative.

(ii) How will Its focal length change In the two media? (CBSE AI 2011)
Answer:
Similarly

Thus f_m = 3.3 f_a, i.e. focal length increases.

Question 2.
A compound microscope uses an objective lens of focal length 4 cm and an eyepiece lens of focal length 10 cm. An object is placed at 6 cm from the objective lens. Calculate the magnifying power of the compound microscope. Also, calculate the length of the microscope. (CBSE Al 2011)
Answer:
f_o = 4 cm, f_e = 10 cm, u_o = – 6 cm, M = ?, L = ?
Using

Hence angular magnification


Question 3.
A giant refracting telescope at an observatory has an objective lens of focal length 15 m. If an eyepiece lens of focal length 1 cm is used, find the angular magnification of the telescope.
If this telescope is used to view the moon, what is the diameter of the image of the moon formed by the objective lens? The diameter of the moon is 3.42 × 10⁶ m and the radius of the lunar orbit is 3.8 × 10⁸ m. (CBSE AI 2011, Delhi 2015)
Answer:
Given f_o = 15 m, f_e = 1.0 cm = 0.01 m, M = ?
Dm = 3.48 × 10⁶ m, r = 3.8 × 10⁸ m,

Using M = f0fe=150.01 = 1500

The angle subtended by the moon at the objective of the telescope

Question 4.
A beam of light converges at a point P. A concave lens of focal length 16 cm is placed in the path of this beam 12 cm from P. Draw a ray diagram and find the location of the point at which the beam would now converge. (CBSE Delhi 2011C)
Answer:
The ray diagram is shown in the figure. In the absence of the concave lens the beam converges at point P. When the concave lens is introduced, the incident beam of light is diverged and now comes to focus at point Q. Thus for the concave lens P serves as a virtual object giving rise to a real Image at Q.

Here u = + 12 cm, f = – 16 cm, v = ?, Now for a lens

Hence v = 48 cm
i. e. the point at which the beam is focused is 48 cm from the lens.

Question 5.
Two convex lenses of focal length 20 cm and 1 cm constitute a telescope. The telescope is focused on a point that is 1 m away from the objective. Calculate the magnification produced and the length of the tube, if the final image Is formed at a distance of 25 cm from the eyepiece.
Answer:
Given f_o = 2o cm, f_e = 1 cm, u = – 100 cm, M =?, y =?

Fora lens 1v−1u=1f
or
1v−1−100=120
or
v = 25cm

Since the eye lens forms the image of the virtual object at the distance of distinct vision for the eye lens
v = – 25 cm, f_e = 1 cm,

Now 1v−1u=1f
or
1−25−1u = 1
or
u = – 2526cm

Now magnification produced by the object lens
m_o = vu=−25100=−14

Magnification produced by the eye Lens
m_e = vu=−25−25 × 26 = 26

Hence total magnification
M = m_o × m_e = -1 /4 × 26 = – 6.5

Question 6.
(a) Under what conditions are the phenomenon of total internal reflection of light observed? Obtain the relation between the critical angle of incidence and the refractive index of the medium.
(b) Three lenses of focal lengths +10 cm, -10 cm and +30 cm are arranged coaxially as in the figure given below. Find the position of the final image formed by the combination.

Answer:
(a) (i) Light travels from a denser medium to a rarer medium.
(ii) Angle of Incidence in the denser medium is more than the critical angle for a given pair of media.
For the grazing incidence n sin i_C = l sin 90°
n = 1sinic

(b) For convex lens f = + 10 cm

Object distance for concave lens u₂ = 15 – 5 = 10 cm

For third lens
1f3=1v3−1∞ ⇒ v₃ = 30 cm

Question 7.
A ray of light incident on an equilateral glass prism (μ_g = 3–√) moves parallel to the baseline of the prism inside it. Find the angle of Incidence for this ray.
Answer:
Given A = 60°, μ_g = 3–√, i = ?

Using the expression μ = sinisinA/2
or

Question 8.
Two monochromatic rays of light are incident normally on the face AB of an isosceles right-angled prism ABC. The refractive indices of the glass prism for the two rays ‘1’ and ‘2’ are respectively 1.35 and 1.45. Trace the path of these rays after entering through the prism.

Answer:
The critical angle for the two rays is

This shows that the angle of Incidence for ray ‘2’ Is greater than the critical angle. Hence it suffers total internal reflection, white ray ‘1’ does not. Hence the path of rays is as shown.

Question 9.
A convex lens of focal length 20 cm is placed coaxially with a convex mirror of radius of curvature 20 cm. The two are kept at 15 cm from each other. A point object lies 60 cm in front of the convex lens. Draw a ray diagram to show the formation of the image by the combination. Determine the nature and position of the image formed. (CBSE AI 2014)
Answer:
The ray diagram is as shown.

For the convex lens, we have
u₁ = – 60 cm, f = + 20 cm, v = ?

Using lens formula we have

Had there been only the Lens, the image would have been formed at Q₁, which acts as a virtual object for the convex mirror.
Therefore u₂ = OQ₁ – OO’ = 30 – 15 = 15 cm

Using mirror formuLa we have
1v2+1u2=2R
or
1v2+1u15=220

Solving for v₂ we have
v₂ = 30cm

Hence the final image is formed at (Point Q) a distance of 30 cm behind the mirror.

Question 10.
A ray PQ is an incident normally on the face AB of a triangular prism refracting angle of 60°, made of a transparent material of refractive index 2 / 3–√, as shown in the figure. Trace the path of the ray as it passes through the prism. Also, calculate the angle of emergence and angle deviation.

Answer:
Critical angle for glass
µ = 1sinic
or
sin i_c = 1μ=3√2= 0.866
or
i_c = 60°

Now the ray is incident at an angle of 60° which is equal to the critical angle, therefore the ray graces the other edge of the prism

Therefore the angle of emergence is = 90°
Hence δ = 30°

In This Post we are  providing Chapter- 8 ELECTROMAGNETIC WAVESNCERT MOST IMPORTANT QUESTIONS for Class 12 PHYSICS which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These MCQS  can be really helpful in the preparation of Board exams and will provide you with a brief knowledge of the chapter

NCERT MOST IMPORTANT QUESTIONS ON ELECTROMAGNETIC WAVES

Question 1.
Answer the following:
(a) Name the em waves which are used for the treatment of certain forms of cancer. Write their frequency range.
Answer:
Gamma rays.
Frequency range > 3 × 10²⁰ Hz

(b) Thin ozone layer on top of the stratosphere is crucial for human survival. Why?
Answer:
The thin ozone layer on top of the stratosphere is crucial for human survival because it absorbs most of the ultraviolet rays coming from the sun. If the ozone layer had not been there, then ultraviolet rays would have entered the earth and caused danger to the survival of the human race.

(c) An em wave exerts pressure on the surface on which it is incident. Justify. (CBSE Delhi 2014)
Answer:
An em wave carries a linear momentum with it. The linear momentum carried by a portion of a wave having energy U is given by p = U/c.

Thus, if the wave incident on a material surface is completely absorbed, it delivers energy U and momentum p = U/c to the surface. If the wave is totally reflected, the momentum delivered is p = 2U/c because the momentum of the wave changes from p to – p. Therefore, it follows that an em wave incident on a surface exerts a force and hence a pressure on the surface.

Question 2.
Answer the following questions:
(a) Why is the thin ozone layer at the top of the stratosphere crucial for human survival? Identify to which part of the electromagnetic spectrum does this radiation belongs and write one important application of the radiation.
Answer:
The thin ozone layer on top of the stratosphere is crucial for human survival because it absorbs most of the ultraviolet rays coming from the sun. If the ozone layer had not been there, then ultraviolet rays would have entered the earth and caused danger to the survival of the human race. This radiation is UV radiation. It is used in sterilization.

(b) Why are infrared waves referred to as heat rays? How are they produced? What role do they play in maintaining the earth’s warmth through the greenhouse effect?
Answer:
Infrared radiations heat up the material on which they fall, hence they are also called heat rays. They are produced by the vibration of atoms and molecules. After falling on the earth, they are reflected back into the earth’s atmosphere. The earth’s atmosphere does not allow these radiations to pass through as such they heat up the earth’s atmosphere.

Question 3.
How are electromagnetic waves produced? What is the source of energy of these waves? Write mathematical expressions for electric and magnetic fields of an electromagnetic wave propagating along the z-axis. Write any two important properties of electromagnetic waves.
Answer:
Electromagnetic waves are produced by accelerated charges which produce an oscillating electric field and magnetic field (which regenerate each other).

• Source of the Energy: Energy of the accelerated charge or the source that accelerates the charges.
• Expression: E_x = E_o sin (kz – ωt) and B_y = B_o sin (kz – ωt)
  (a) They are transverse in nature.
  (b) They don’t require a medium to propagate.

Question 4.
How are em waves produced by oscillating charges?
Draw a sketch of linearly polarised em waves propagating in the Z-direction. Indicate the directions of the oscillating electric and magnetic fields.
Answer:
(a) An oscillating charge produces an oscillating electric field in space, which produces an oscillating magnetic field. The oscillating electric and magnetic fields regenerate each other, and this results in the production of em waves in space.
(b) See Figure.

Question 5.
Write Maxwell’s generalization of Ampere’s Circuital Law. Show that in the process of charging a capacitor, the current produced within the plates of the capacitor is i = ε_odϕEdt where ΦE is the electric flux produced during charging of the capacitor plates.
Answer:
The generalized form of Maxwell ampere law is
∮B⃗ ⋅dl→= μ_o(l + l_D) where l_D = ε_odϕEdt=dqdt

The electric flux Φ between the plates of the parallel plate capacitor through which a time-dependent current flow is given by:
Φ_E = E A, but E = σ/ε_o
Therefore we have

Question 6.
(a) Why are Infrared waves often called heatwaves? Explain.
Answer:
Infrared waves have frequencies lower than those of visibLe Light; they have the ability to vibrate not only the electrons but the entire atoms or molecules of a body. This vibration increases the internal energy and temperature of the body. That is why infrared waves are often called heat waves.

(b) What do you understand by the statement, “Electromagnetic waves transport momentum”? (CBSE AI, Delhi 2018)
Answer:
If we consider a plane perpendicular to the direction of propagation of the electromagnetic wave, then electric charges present on the plane will be set and sustained in motion by the electric and magnetic fields of the electromagnetic wave. The charges present on the surface thus acquire energy and momentum from the waves. This just illustrates the fact that an electromagnetic wave (like other waves) transfers energy and momentum.

Question 7.
(a) When the oscillating electric and magnetic fields are along the x- and y-direction respectively
(i) point out the direction of propagation of the electromagnetic wave,
Answer:
Z-axis

(ii) express the velocity of propagation in terms of the amplitudes of the oscillating electric and magnetic fields.
Answer:
c = E_o / B_o

(b) How do you show that the em wave carries energy and momentum? (CBSE A! 2013C)
Answer:
Consider a plane perpendicular to the direction of propagation of the electromagnetic wave. If there are, on this plane, electric charges, they will be set and sustained in motion by the electric and magnetic fields of the electromagnetic wave. The charges thus acquire energy and momentum from the waves. This illustrates the fact that an electromagnetic wave carries energy and momentum.

Question 8.
A radio can tune In to any station in the 7.5 MHz to 12 MHz bands. What is the corresponding wavelength band? (NCERT)
Answer:

Thus the wavelength band is 40 m to 25m.

Question 9.
Suppose that the electric field amplitude of an electromagnetic wave E₀ = 120 N C^-1 and that Its frequency is v = 50.0 MHz. (a) Determine, B0, ω k, and λ. (b) Find expressions for E and B. (NCERT)
Answer:


Question 10.
In a plane electromagnetic wave, the electric field oscillates sinusoidally at a frequency of 2.0 × 10¹⁰ Hz and amplitude 48 V m^-1.
(a) What is the wavelength of the wave?
(b) What is the amplitude of the oscillating magnetic field?
(c) Show that the average energy density of the E field equals the average energy density of the 8 fields. [c = 3 × 10⁸ms^-1.]
Answer:

(c) The average density of electric field is
given by U_e = 12ε₀E² and the average energy density of the magnetic field is given by U_B = B22μ0. But B = Ec and C = 1μ0ε0√ , hence the above equation becomes UB = B22μ0=E22μ0c2,