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NCERT solutions for class 12 Physics 2022 - While doing homework issues, having a tool like NCERT Class 12 Physics chapter 5 magnetism and matter solutions in your hand would be beneficial. You learned how to obtain isolated charge in the first chapter. Is it possible to get a magnetic monopole? When a magnet is divided into numerous parts, each portion acts as a magnet with a south and north pole, or magnetic dipole. For latest updates on NCERT Solutions for class 12 Physics 2022, read this article.

by Keerthika | Last updated: Feb 01, 2022

NCERT - Learn Chapter-wise Free NCERT Text books

NCERT - Learn Chapter-wise Free NCERT Text books

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NCERT solutions for class 12 Physics 2022

New Update on 01.02.2022: NCERT solutions for class 12 Physics 2022 - When doing homework issues, having a tool like NCERT class 12 physics chapter 5 magnetism and matter solutions in your hand would be beneficial. Isolated charge can be obtained, as you learned in the first chapter. Is it possible to obtain a magnetic monopole? When a magnet is cut into numerous pieces, each piece acts as a magnet with a south and north pole, or magnetic dipole.

This proves that there is no such thing as a magnetic monopole. You'll also learn about magnet characteristics in magnetism and matter. The NCERT class 12 Physics chapter 5 solutions include questions on all of the chapter's subjects. You can compare this chapter to electrostatics when studying it.

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Chapters In NCERT Class 12 Physics 2022

  • Chapter 1 - Electric Charges and Fields

  • Chapter 2 - Electrostatic Potential and Capacitance

  • Chapter 3 - Current Electricity

  • Chapter 4 - Moving Charges and Magnetism

  • Chapter 5 - Magnetism and Matter

  • Chapter 6 - Electromagnetic Induction

  • Chapter 7 - Alternating Current

  • Chapter 8 - Electromagnetic Waves

  • Chapter 9 - Ray Optics and Optical Instruments

  • Chapter 10 - Wave Optics Solutions

  • Chapter 11 - Dual nature of radiation and matter

  • Chapter 12 - Atoms

  • Chapter 13 - Nuclei

  • Chapter 14 - Semiconductor Electronics Materials Devices and Simple Circuits

Chapter Wise NCERT Solutions For Class 12 Physics 2022

Chapter 1 - Electric Charges and Fields

1. Explain the meaning of the statement ‘electric charge of a body is quantised’.

Answer:

The given statement "electric charge of a body is quantised" implies that charge on a body can take only integral values. In other words, only the integral number of electrons can be transferred from one body to another and not in fractions.

Therefore, a charged body can only have an integral multiple of the electric charge of an electron.

2. Why can one ignore quantisation of electric charge when dealing with macroscopic i.e., large scale charges?

Answer:

On a macroscopic level, the amount of charge transferred is very large as compared to the charge of a single electron. Therefore, we tend to ignore the quantisation of electric charge in these cases and consider it to be continuous in nature.

3. When a glass rod is rubbed with a silk cloth, charges appear on both. A similar phenomenon is observed with many other pairs of bodies. Explain how this observation is consistent with the law of conservation of charge.

Answer:

When a glass rod is rubbed with a silk cloth, opposite charges appear on both the rod and the cloth.

The phenomenon of charging bodies by rubbing them against each other is known as charging by friction. Here, electrons are transferred from one body to another giving both the bodies an equal but opposite charge. The number of electrons lost by one body (attains positive charge due to loss of negatively charged electrons) is equal to the number of electrons gained by the other body (attains negative charge). Therefore, the net charge of the system is zero. This is in accordance with the law of conservation of charge

4. An electrostatic field line is a continuous curve. That is, a field line cannot have sudden breaks. Why not?

Answer:

A positive point charge experiences a force in an electrostatic field. Since the charge will experience a continuous force and cannot jump from one point to another, the electric field lines must be continuous.

5. Explain why two field lines never cross each other at any point?

Answer:

A tangent drawn at any point on a field line gives the direction of force experienced by a unit positive charge due to the electric field on that point. If two lines intersect at a point, then the tangent drawn there will give two directions of force, which is not possible. Hence two field lines cannot cross each other at any point.

6. A sensitive instrument is to be shielded from the strong electrostatic fields in its environment. Suggest a possible way.

Answer:

We know that the electric field inside a conductor is zero.

Therefore, a possible way to shield from the strong electrostatic fields in its environment is to enclose the instrument fully by a metallic surface.

7. Verify this result for the simple configuration of two charges of the same magnitude and sign placed a certain distance apart.

Answer:

Two charges of the same magnitude and sign are placed at a certain distance apart. The mid-point of the line joining these charges will have E =0.

When a test charge is displaced along the line towards the charges, it experiences a restoring force(which is the condition for stable equilibrium). But if the test charge is displaced along the normal of the line, the net electrostatic force pushes it away from the starting point. Hence, the equilibrium is unstable.

Chapter 2 - Electrostatic Potential and Capacitance

1. A spherical conductor of radius 12 cm has a charge of 1.6 times 10^{-7}C distributed uniformly on its surface. What is the electric field (a) inside the sphere

Answer:

Since the charge is uniformly distributed and it always remains on the surface of the conductor, the electric field inside the sphere will be zero.

2. A spherical conducting shell of inner radius r 1 and outer radius r 2 has a charge Q.

(b)Is the electric field inside a cavity (with no charge) zero, even if the shell is not spherical, but has any irregular shape? Explain.

Answer:

Yes, the electric field inside the cavity is zero even when the shape is irregular and not the sphere. Suppose a Gaussian surface inside the cavity, now since there is no charge inside it, the electric flux through it will be zero according to the guess law. Also, all of the charges will reside on the surface of the conductor so, net charge inside is zero. hence electric field inside cavity is zero.

3. What meaning would you give to the capacitance of a single conductor?

Answer:

There is no meaning in the capacitor with a single plate factually. but we give it meaning by assuming the second plate at infinity. Hence capacitance of a single conductor is the amount of change required to raise the potential of the conductor by one unit amount.

4. Two charges –q and +q are located at points (0, 0, –a) and (0, 0, a), respectively

(c) How much work is done in moving a small test charge from the point (5,0,0) to (–7,0,0) along the x-axis? Does the answer change if the path of the test charge between the same points is not along the x-axis?

Answer:

Since point (5,0,0) is equidistance from both charges, they both will cancel out each other potential and hence potential at this point is zero.

Similarly, point (–7,0,0) is also equidistance from both charges. and hence potential at this point is zero.

Since potential at both the point is zero, the work done in moving charge from one point to other is zero. Work done is independent of the path.

5. What is the work done by the field of a nucleus in a complete circular orbit of the electron? What if the orbit is elliptical?

Answer:

The initial and final position will be the same for any orbit whether it is circular or elliptical. Hence work done will always be zero.

6. We know that electric field is discontinuous across the surface of a charged conductor. Is electric potential also discontinuous there?

Answer:

Since the electric potential is not a vector quantity unlike the electric field, it can never be discontinuous.

7. Guess a possible reason why water has a much greater dielectric constant (= 80) than say, mica (= 6).

Answer:

Water has a much greater dielectric constant than mica because it posses a permanent dipole moment and has an unsymmetrical shape.

Chapter 3 - Current Electricity

1. The storage battery of a car has an emf of 12 V. If the internal resistance of the battery is 0.4 Omega , what is the maximum current that can be drawn from the battery?

Answer :

Given, the emf of battery, E = 12 V

Internal resistance of battery, r = 0.4 Ohm

Let I be the maximum current drawn from the battery.

We know, according to Ohm's law

E = Ir

I = E/r = 12/0.4 =30 A

Hence the maximum current drawn from the battery is 30 A.

2. A steady current flows in a metallic conductor of non-uniform cross-section. Which of these quantities is constant along the conductor: current, current density, electric field, drift speed?

Answer:

The current flowing through the conductor is constant for a steady current flow.

Also, current density, electric field, and drift speed are inversely proportional to the area of cross-section. Hence, not constant.

3. Is Ohm’s law universally applicable for all conducting elements? If not, give examples of elements which do not obey Ohm’s law.

Answer:

No. Ohm’s law is not universally applicable for all conducting elements.

A semiconductor diode is such an example.

4. A low voltage supply from which one needs high currents must have very low internal resistance. Why?

Answer:

Ohm's law states that: V = I x R

Hence for a low voltage V, resistance R must be very low for a high value of current.

5. A high tension (HT) supply of, say, 6 kV must have a very large internal resistance. Why?

Answer:

A very high internal resistance is required for a high tension supply to limit the current drawn for safety purposes.

6. Given n resistors each of resistance R , how will you combine them to get the (i) maximum effective resistance?

Answer:

To get maximum effective resistance, combine them in series. The effective resistance will be nR.

7. Given n resistors each of resistance R, how will you combine them to get the (ii) minimum effective resistance?

Answer:

To get minimum effective resistance, combine them in parallel. The effective resistance will be R/n.

Chapter 4 - Moving Charges and Magnetism

1. A toroid has a core (non-ferromagnetic) of inner radius 25 cm and outer radius 26 cm, around which 3500 turns of a wire are wound. If the current in the wire is 11 A, what is the magnetic field in the empty space surrounded by the toroid?

Answer:

The magnetic field in the empty space surrounded by the toroid is zero.

2. A magnetic field that varies in magnitude from point to point but has a constant direction (east to west) is set up in a chamber. A charged particle enters the chamber and travels undeflected along a straight path with constant speed. What can you say about the initial velocity of the particle?

Answer:

The charged particle is not deflected by the magnetic field even while having a non zero velocity, therefore, its initial velocity must be either parallel or anti-parallel to the magnetic field i.e. It's velocity is either towards the east or the west direction.

3. A charged particle enters an environment of a strong and non-uniform magnetic field varying from point to point both in magnitude and direction, and comes out of it following a complicated trajectory. Would its final speed equal the initial speed if it suffered no collisions with the environment?

Answer:

Yes, its final speed will be equal to the initial speed if it has not undergone any collision as the work done by the magnetic field on a charged particle is always zero because it acts perpendicular to the velocity of the particle.
4. An electron traveling west to east enters a chamber having a uniform electrostatic field in north to south direction. Specify the direction in which a uniform magnetic field should be set up to prevent the electron from deflecting from its straight line path.

Answer:

The electron would experience an electrostatic force towards the north direction, therefore, to nullify its force due to the magnetic field must be acting on the electron towards the south direction. By using Fleming's left-hand rule we can see that the force will be in the north direction if the magnetic field is in the vertically downward direction.

5. A circular coil of 20 turns and radius 10 cm is placed in a uniform magnetic field of 0.10 T normal to the plane of the coil. If the current in the coil is 5.0 A, what is the total torque on the coil,

Answer:

As we know the torque on a current-carrying loop in a magnetic field is given by the following relation

vec{tau }=Ivec{A}times vec{B}

It is clear that the torque, in this case, will be 0 as the area vector is along the magnetic field only.
6. A circular coil of 20 turns and radius 10 cm is placed in a uniform magnetic field of 0.10 T normal to the plane of the coil. If the current in the coil is 5.0 A, total force on the coil.

Answer:

The total force on the coil will be zero as the magnetic field is uniform

7. A square coil of side 10 cm consists of 20 turns and carries a current of 12 A. The coil is suspended vertically and the normal to the plane of the coil makes an angle of 30 degree with the direction of a uniform horizontal magnetic field of magnitude 0.80 T. What is the magnitude of torque experienced by the coil?

Answer:

The magnitude of torque experienced by a current-carrying coil in a magnetic field is given by

tau =nBIAsintheta

where n = number of turns, I is the current in the coil, A is the area of the coil and theta is the angle between the magnetic field and the vector normal to the plane of the coil.

In the given question n = 20, B=0.8 T, A=0.1 times 0.1=0.01 m 2 , I=12 A, theta =30 o

tau =20times 0.8times 12times 0.01times sin30^{o}

=0.96 Nm

The coil, therefore, experiences a torque of magnitude 0.96 Nm.

Chapter 5 - Magnetism and Matter

1. Answer the following questions regarding earth’s magnetism: A vector needs three quantities for its specification. Name the three independent quantities conventionally used to specify the earth’s magnetic field.

Answer:

The three independent quantities used to specify the earth’s magnetic field are:

(i) The horizontal component of Earth's magnetic field ( H_{E} ).

(ii) The magnetic declination (D): It is the angle between the geographic north and the magnetic north at a place.

(iii)The magnetic dip (I): It is the angle between the horizontal plane and the magnetic axis, as observed in the compass


2. Answer the following questions regarding earth’s magnetism. The angle of dip at a location in southern India is about 18°. Would you expect a greater or smaller dip angle in Britain?

Answer:

We would expect a greater angle of dip in Britain. The angle of dip increases as the distance from equator increases.

(It is 0 at the equator and 90 degrees at the poles)

3. If you made a map of magnetic field lines at Melbourne in Australia, would the lines seem to go into the ground or come out of the ground?

Answer:

The field lines go into the earth at the north magnetic pole and come out from the south magnetic pole and hence Australia being in the southern hemisphere. The magnetic field lines would come out of the ground at Melbourne.

4. In which direction would a compass free to move in the vertical plane point to, if located right on the geomagnetic north or south pole?

Answer:

The magnetic field is perpendicular at the poles and the magnetic needle of the compass tends to align with the magnetic field. Therefore the compass will get aligned in the vertical direction if is held vertically at the north pole.

5. Geologists claim that besides the main magnetic N-S poles, there are several local poles on the earth’s surface oriented in different directions. How is such a thing possible at all?

Answer:

This may be possible due to the presence of minerals which are magnetic in nature.

6. The earth’s magnetic field varies from point to point in space. Does it also change with time? If so, on what time scale does it change appreciably?

Answer:

Due to the constant but slow motion of the plates and change in the core, magnetic field due to Earth may change with time too. The time scale is in centuries for appreciable change.

7. The earth’s core is known to contain iron. Yet geologists do not regard this as a source of the earth’s magnetism. Why?

Answer:

The iron present in the core of the Earth is in the molten form. Hence it loses it ferromagnetism and not regarded by geologists as a source of earth's magnetism.

Chapter 6 - Electromagnetic Induction

1. A horizontal straight wire 10m long extending from east to west is falling with a speed of 5.0 m: s^{-1} , at right angles to the horizontal component of the earth’s magnetic field, 0.30times 10^{-4}: wb: m^{-2} . What is the instantaneous value of the emf induced in the wire?

Answer:

Given

Length of the wire l=10m

Speed of the wire v=5m/s

The magnetic field of the earth B=0.3*10^{-4}Wbm^{-2}

Now,

The instantaneous value of induced emf :

e=Blv=0.3*10^{-4}*10*5=1.5*10^{-3}

Hence instantaneous emf induce is 1.5*10^{-3} .

2. A horizontal straight wire 10m long extending from east to west is falling with a speed of 5.0 m: s^{-1} , at right angles to the horizontal component of the earth’s magnetic field, 0.30times 10^{-4}: wb: m^{-2} . What is the direction of the emf?

Answer:

If we apply the Flemings right-hand rule, we see that the direction of induced emf is from west to east.

3. Use Lenz’s law to determine the direction of induced current

Answer

By turning the wire from irregular shape to circle, we are increasing the area of the loop so flux will increase so current will induce in such a way that reduces the flux through it. By right-hand thumb rule direction of current is adcba.

4. An air-cored solenoid with length 30cm , area of cross-section 25cm^{2} and number of turns 500 , carries a current of 2.5A . The current is suddenly switched off in a brief time of 10^{-3}s . How much is the average back emf induced across the ends of the open switch in the circuit? Ignore the variation in magnetic field near the ends of the solenoid.

Answer:

Given

Length od the solenoid l=30cm=0.3m

Area of the cross-section of the solenoid A=25cm^2=25*10^{-4}m^2

Number of turns in the solenoid N=500

Current flowing in the solenoid I=2.5A

The time interval for which current flows Delta t=10^{-3}s

5. A metal rod PQ resting on the smooth rails AB and positioned between the poles of a permanent magnet. The rails, the rod, and the magnetic field are in three mutual perpendicular directions. A galvanometer G connects the rails through a switch K . Length of the rod=15cm , B=0.50T , resistance of the closed loop containing the rod=9.0mOmega . Assume the field to be uniform. Suppose K is open and the rod is moved with a speed of 12cm: s^{-1} in the direction shown. Give the polarity and magnitude of the induced emf
What is the induced emf in the moving rod if the magnetic field is parallel to the rails instead of being perpendicular?

Answer

If the magnetic field is parallel to the rail then, the motion of the rod will not cut across the magnetic field lines and hence no emf will induce. Hence emf induced is zero in this case.

Chapter 7 - Alternating Current

1. A 100Omega resistor is connected to a 220: V , 50: Hz ac supply. What is the net power consumed over a full cycle?

Answer:

Given,

Supplied RMS Voltage V_{rms}=220V

Supplied RMS Current I_{rms}=2.2A

The net power consumed over a full cycle:

P=V_{rms}I_{rms}=220*2.2=484W

Hence net power consumed is 484W.

2. A circuit containing a 80mH inductor and a 60mu F capacitor in series is connected to a 230: V , 50: Hz supply. The resistance of the circuit is negligible . What is the total average power absorbed by the circuit? [‘Average’ implies ‘averaged over one cycle’.]

Answer:

Since the phase difference between voltage and current is 90 degree, even the total power absorbed by the circuit is zero. This is an ideal circuit, we can not have any circuit in practical that consumes no power, that is because practically resistance of any circuit is never zero. Here only inductor and capacitor are present and none of them consumes energy, they just store it and transfer it like they are doing it in this case.

3. What is the source frequency for which average power absorbed by the circuit is maximum. Obtain the value of this maximum power.

Answer:

Since the resistor is the only element in the circuit which consumes the power, the maximum absorbed power by circuit will be maximum when power absorbed by the resistor will be maximum. power absorbed by the resistor will be maximum at when current is maximum which is the natural frequency case,

Hence when source frequency will be equal to the natural frequency, the power absorbed will be maximum.

4. In any ac circuit, is the applied instantaneous voltage equal to the algebraic sum of the instantaneous voltages across the series elements of the circuit? Is the same true for rms voltage?

Answer:

Yes, at any instant the applied voltage will be distributed among all element and the sum of the instantaneous voltage of all elements will be equal to the applied. But this is not the case in RMS because all elements are varying differently and they may not be in the phase.

5. A capacitor is used in the primary circuit of an induction coil.

Answer:

Yes, we use capacitors in the primary circuit of an induction coil to avoid sparking. when the circuit breaks, a large emf is induced and the capacitor gets charged from this avoiding the case of sparking and short circuit.

6. A choke coil in series with a lamp is connected to a dc line. The lamp is seen to shine brightly. Insertion of an iron core in the choke causes no change in the lamp’s brightness. Predict the corresponding observations if the connection is to an ac line.

Answer:

For a steady state DC, the increasing inductance value by inserting iron core in the choke, have no effect on the brightness of the connected lamp, whereas, for ac when the iron core is inserted, the light of the lamp will shine less brightly because of increase in inductive impedance.

7. Why is choke coil needed in the use of fluorescent tubes with ac mains? Why can we not use an ordinary resistor instead of the choke coil?

Answer:

We need a choke coil in the use of fluorescent tubes with ac mains to reduce the voltage across the tube without wasting much power. If we use a simple resistor for this purpose, there will be more power loss, hence we do not prefer it.

Chapter 8 - Electromagnetic Waves

1. A capacitor made of two circular plates each of radius 12 cm, and separated by 5.0 cm. The capacitor is being charged by an external source (not shown in the figure). The charging current is constant and equal to 0.15A. Is Kirchhoff’s first rule (junction rule) valid at each plate of the

capacitor? Explain.

Answer:

Yes, Kirchoff's First rule (junction rule) is valid at each plate of the capacitor. This might not seem like the case at first but once we take into consideration both the conduction and displacement current the Kirchoff's first rule will hold good.

2. The small ozone layer on top of the stratosphere is crucial for human survival. Why?

Answer:

The small ozone layer on top of the stratosphere is crucial for human survival as it absorbs the ultraviolet radiations coming from the sun which are very harmful to humans.

3. A charged particle oscillates about its mean equilibrium position with a frequency of 10^{9} Hz. What is the frequency of the electromagnetic waves produced by the oscillator?

Answer:

The frequency of the electromagnetic waves produced by the oscillation of a charged particle about a mean position is equal to the frequency of the oscillation of the charged particle. Therefore electromagnetic waves produced will have a frequency of 10 9 Hz.

4. State the part of the electromagnetic spectrum to which each belongs. 21 cm (wavelength emitted by atomic hydrogen in interstellar space).

Answer:

Radio waves

5. Long distance radio broadcasts use short-wave bands. Why?

Answer:

Long-distance radio broadcasts use short-wave bands as these are refracted by the ionosphere.

6. It is necessary to use satellites for long distance TV transmission. Why?

Answer:

As TV signals are of high frequencies they are not reflected by the ionosphere and therefore satellites are to be used to reflect the

7. Optical and radio telescopes are built on the ground but X-ray astronomy is possible only from satellites orbiting the earth. Why?

Answer:

X-rays are absorbed by the atmosphere and therefore the source of X-rays must lie outside the atmosphere to carry out X-ray astronomy and therefore satellites orbiting the earth are necessary but radio waves and visible light can penetrate through the atmosphere and therefore optical and radio telescopes can be built on the ground.

Chapter 9 - Ray Optics and Optical Instruments

1. You have learned that plane and convex mirrors produce virtual images of objects. Can they produce real images under some circumstances? Explain.

Answer:

If our object is virtual then Plane and convex mirrors can produce a real image. That is, when the light coming from infinity goes into the convex mirror, it creates a virtual object behind the convex mirror. the reflection of this virtual object in the convex mirror can be taken out on screen and hence convex mirror can make a real image.

2. A virtual image, we always say, cannot be caught on a screen. Yet when we ‘see’ a virtual image, we are obviously bringing it on to the ‘screen’ (i.e., the retina) of our eye. Is there a contradiction?

Answer:

No, there is no contradiction. A virtual image is formed whenever the light rays are diverging. We have a convex lens in our eye. This convex lens converges the diverging rays into our retina and forms a real image. In other words, the virtual image acts as an object to the convex lens of our eye to form a real image, which we see on the screen called retina.

3. A diver underwater looks obliquely at a fisherman standing on the bank of a lake. Would the fisherman look taller or shorter to the diver than what he actually is?

Answer:

The diver is in denser medium (water) and fisherman is in lighter medium (air). As the diver is looking at the fisherman, rays of light will go from fisherman to divers eye, that is, from lighter medium to denser medium. Since rays deflect toward normal when it goes form lighter to a denser medium, the fisherman will look taller than actual to the diver.

4. Does the apparent depth of a tank of water change if viewed obliquely? If so, does the apparent depth increase or decrease?

Answer:

Yes, appearing depth of water will decrease when we view obliquely, this happens because of the fact that light bends from its direction whenever it goes from one medium to another medium.

5. The refractive index of diamond is much greater than that of ordinary glass. Is this fact of some use to a diamond cutter?

Answer:

We use diamond as a cutter because it is very hard and sharp. The refractive index is high in diamond ensures that light goes through multiple total internal reflections so that light goes in all direction. This is the reason behind the shining of the diamond. Light entering is totally reflected from faces before it getting out, hence producing a sparkling effect


6. The angle subtended at the eye by an object is equal to the angle subtended at the eye by the virtual image produced by a magnifying glass. In what sense then does a magnifying glass provide angular magnification?

Answer:

Angular magnification is the ratio of tangents of the angle formed by object and image from the centre point of the lens. In this question angle formed by the object and a virtual image is same but it provides magnification in a way that, whenever we have object place before 25cm, the lens magnifies it and make it in the vision range. By using magnification we can put the object closer to the eye and still can see it which we couldn't have without magnification.

7. In viewing through a magnifying glass, one usually positions one’s eyes very close to the lens. Does angular magnification change if the eye is moved back?

Answer:

Yes, angular magnification will change if we move our eye away from the lens. this is because then angle subtended by lens would be different than the angle subtended by eye. When we move our eye form lens, angular magnification decreases. Also, one more important point here is that object distance does not have any effect on angular magnification.

Chapter 10 - Wave Optics Solutions

1. Light emerging out of a convex lens when a point source is placed at its focus.

Answer:

The shape of light emerging out of a convex lens when a point source is placed at its focus is Parallel .when light rays come from infinity parallelly, they intersect at focus of convex lens and hence when light is emerging from the focus, the rays will get parallel to each other after coming out of the convex lens, because path of light rays are reversible.

2. What is the shape of the wavefront in each of the following cases: The portion of the wavefront of light from a distant star intercepted by the Earth

Answer:

The portion of the wavefront of the light from the distant star which is seen from earth is plane since a small area of a large sphere will nearly look like a plane.

3. Let us list some of the factors, which could possibly influence the

speed of wave propagation:

(i) nature of the source.

ii) direction of propagation.

(iii) motion of the source and/or observer.

(iv) wavelength.

(v) intensity of the wave.

On which of these factors, if any, does the speed of light in vacuum depend?

Answer:

The speed of light in a vacuum is constant and independent of anything according to Einstein's theory of relativity.

4. Let us list some of the factors, which could possibly influence the speed of wave propagation:

(i) nature of the source.

(ii) direction of propagation.

(iii) motion of the source and/or observer.

(iv) wavelength.

(v) intensity of the wave.

On which of these factors, if any, does the speed of light in a medium (say, glass or water), depend?

Answer:

The speed of light in any medium depends upon the wavelength of the light and does not depends on the nature of the source, direction of propagation, the motion of the source and/or observer, and intensity of the wave.

5. For sound waves, the Doppler formula for frequency shift differs slightly between the two situations: (i) source at rest; observer moving, and (ii) source moving; observer at rest. The exact Doppler formulas for the case of light waves in vacuum are, however, strictly identical for these situations. Explain why this should be so. Would you expect the formulas to be strictly identical for the two situations in case of light travelling in a medium?

Answer:

The sound wave requires a medium for propagation.so, even though both given situation may relate to the same relative motion, they are not identical physically since, the motion of the observer, relative to the medium is different in two situations. Hence, we cannot expect the Doppler formula to be identical in both given cases.

When light waves are in a vacuum, there is clearly nothing to distinguish between two cases. for light propagation in a medium, two situations are not identical for the same reason as in the case of sound waves.

6. In what way is diffraction from each slit related to the interference pattern in a double-slit experiment?

Answer:

When we have a width in the order of lambda , the intensity of interference fringes in Young's double-slit experiment is modified by the diffraction pattern of each slit.

7. 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?

Answer:

A bright spot is seen at the centre of the shadow of the obstacle because wave diffracted from the edge of a circular obstacle interfere constructively at the centre of the shadow producing the bright spot.

Chapter 11 - Dual nature of radiation and matter

1. The work function of caesium metal is 2.14hspace{1mm}eV When light of frequency 6times 10^1^4hspace{1mm}Hz is incident on the metal surface, photoemission of electrons occurs. What is the stopping potential

Answer:

The stopping potential depends on the maximum Kinetic Energy of the emitted electrons. Since maximum Kinetic energy is equal to 0.34 eV, stopping potential is the maximum kinetic energy by charge equal to 0.34 V.

2. Every metal has a definite work function. Why do all photoelectrons not come out with the same energy if incident radiation is monochromatic? Why is there an energy distribution of photoelectrons?

Answer:

The work function is defined as the minimum energy below which an electron will never be ejected from the metal. But when photons with high energy are incident it is possible that electrons from different orbits get ejected and would, therefore, come out of the atom with different kinetic energies.

3. Why should gases be insulators at ordinary pressures and start conducting at very low pressures?

Answer:

At ordinary pressure due to a large number of collisions among themselves, the gases have no chance of reaching the electrodes while at very low pressure these collisions decrease exponentially and the gas molecules have a chance of reaching the respective electrodes and therefore are capable of conducting electricity.

4. Quarks inside protons and neutrons are thought to carry fractional charges dpi{100} [(+2/3)e;(-1/3)e] . Why do they not show up in Millikan’s oil-drop experiment?

Answer:

Quarks are thought to be tight within a proton or neutron by forces which grow tough if one tries to pull them apart. That is event though fractional charges may exist in nature, the observable charges are still integral multiples of the charge of the electron

Chapter 12 - Atoms

1. Suppose you are given a chance to repeat the alpha-particle scattering experiment using a thin sheet of solid hydrogen in place of the gold foil. (Hydrogen is a solid at temperatures below 14 K.) What results do you expect?

Answer:

On repeating the alpha-particle scattering experiment using a thin sheet of solid hydrogen in place of the gold foil we would have different observations than Rutherford, as the alpha particles won't be scattered much because of being heavier than the nucleus of the Hydrogen atom. Therefore we would not be able to confirm the presence of almost the entire mass of the atom at its centre.
2. Which help you understand the difference between Thomson’s model and Rutherford’s model better.

Is the average angle of deflection of alpha -particles by a thin gold foil predicted by Thomson’s model much less, about the same, or much greater than that predicted by Rutherford’s model?

Answer:

The average angle of deflection of alpha -particles by a thin gold foil predicted by both the models is about the same.

3. Which help you understand the difference between Thomson’s model and Rutherford’s model better.

Is the probability of backward scattering (i.e., scattering of alpha -particles at angles greater than 90^{circ} ) predicted by Thomson’s model much less, about the same, or much greater than that predicted by Rutherford’s model?

Answer:

The probability of backward scattering predicted by Thomson’s model is much less than that predicted by Rutherford’s model

4. Which help you understand the difference between Thomson’s model and Rutherford’s model better.

Keeping other factors fixed, it is found experimentally that for small thickness t, the number of alpha -particles scattered at moderate angles is proportional to t. What clue does this linear dependence on t provide?

Answer:

Scattering at moderate angles requires head-on collision the probability of which increases with the number of target atoms in the path of alpha -particles which increases linearly with the thickness of the gold foil and therefore the linear dependence between the number of alpha -particles scattered at a moderate angle and the thickness t of the gold foil.

5. Which help you understand the difference between Thomson’s model and Rutherford’s model better.

In which model is it completely wrong to ignore multiple scattering for the calculation of average angle of scattering of alpha -particles by a thin foil?

Answer:

It is completely wrong to ignore multiple scattering for the calculation of the average angle of scattering of alpha -particles by a thin foil in Thomson's model as the deflection caused by a single collision in this model is very small.

Chapter 13 - Nuclei

1. Obtain the binding energy( in MeV ) of a nitrogen nucleus (_{7}^{14}textrm{N}) , given m (_{7}^{14}textrm{N})=14.00307; ; u

Answer:

m n = 1.00866 u

m p = 1.00727 u

Atomic mass of Nitrogen m= 14.00307 u

Mass defect Delta m=7 times m n +7 times m p - m

Delta m=7 times 1.00866+7 times 1.00727 - 14.00307

Delta m=0.10844

Now 1u is equivalent to 931.5 MeV

E b =0.10844 times 931.5

E b =101.01186 MeV

Therefore binding energy of a Nitrogen nucleus is 101.01186 MeV.

2. Write nuclear reaction equations for

(i); alpha -decay; of; _{88}^{226}textrm{Ra}

Answer:

The nuclear reaction equations for the given alpha decay

_{88}^{226}textrm{Ra}rightarrow _{86}^{222}textrm{Rn}+_{2}^{4}textrm{He}

3. Write nuclear reaction equations for

(ii); alpha -decay; of; _{94}^{242}textrm{Pu}

Answer:

The nuclear reaction equations for the given alpha decay is

_{94}^{242}textrm{Pu}rightarrow _{92}^{238}textrm{U}+_{2}^{4}textrm{He}

4. Write nuclear reaction equations for

(v); beta ^{+} -: decay; of; _{6}^{11}textrm{C}

Answer:

The nuclear reaction for the given beta plus decay will be

_{6}^{11}textrm{C}rightarrow _{5}^{11}textrm{P}+e^{+}+nu

Chapter 14 - Semiconductor Electronics Materials Devices and Simple Circuits

1. In an unbiased p-n junction, holes diffuse from the p-region to n-region beca

(a) free electrons in the n-region attract them

(b) they move across the junction by the potential difference.

(c) hole concentration in p-region is more as compared to n-region.

d) All the above

Answer:

Charge flows from higher concentration to the lower concentration in a junction. In this case, holes are diffusing from the p-region to n-region and hence the concentration of hole is greater in p region.

and hence correct option would be (c)

2. When a forward bias is applied to a p-n junction, it

(a) raises the potential barrier

(b) reduces the majority carrier current to zero.

(c) lowers the potential barrier.

(d) none of the above.

Answer:

When a p-n junction is forward biased, the negative voltage repels the electron toward junction and give them the energy to cross the junction and combine with the hole which is also being pushed by a positive voltage. This leads to a reduction in the depletion layer which means a reduction in potential barrier across the junction.

Hence correct option would be (c)

3. In half-wave rectification, what is the output frequency if the input frequency is 50 ; Hz. What is the output frequency of a full-wave rectifier for the same input frequency

Answer:

As we know : output frequency for half-wave rectifier = input frequency, and hence output frequency in half-wave rectifier will be 50Hz.

4. In an n-type silicon, which of the following statement is true:

(a) Electrons are majority carriers and trivalent atoms are the dopants.

(b) Electrons are minority carriers and pentavalent atoms are the dopants.

(c) Holes are minority carriers and pentavalent atoms are the dopants.

(d) Holes are majority carriers and trivalent atoms are the dopants.

Answer:

An N-type semiconductor has electron as majority carriers and holes as minority carriers. It is formed when we dope pentavalent impurity in Silicon atom. Some pentavalent dopants are phosphorus, arsenic, and bismuth.

Hence the correct option is C
5. Which of the statements given in Exercise 14.1 is true for p-type semiconductors.

(a) Electrons are majority carriers and trivalent atoms are the dopants.

(b) Electrons are minority carriers and pentavalent atoms are the dopants.

(c) Holes are minority carriers and pentavalent atoms are the dopants.

(d) Holes are majority carriers and trivalent atoms are the dopants

Answer:

In a p-type semiconductor, holes are the majority carrier and electrons are the minority carrier. It is formed when a trivalent atom-like aluminium is doped in a silicon atom. Hence correct option for p-type conductor would be (d).

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Also Learn

Ncert solutions 2023 - FAQ

1. What is the first chapter for Class 12 NCERT?

Electric Charges and Fields  is the first chapter for Class 12 NCERT

2. What is the 8th chapter for Class 12 NCERT?

Electromagnetic Waves  is the 8th chapter for Class 12 NCERT

3. What is the 14th chapter for Class 12 NCERT?

Semiconductor Electronics Materials Devices and Simple Circuits  is the 14th chapter for Class 12 NCERT

4. What is the 6th chapter for Class 12 NCERT?

Electromagnetic Induction  is the 6th chapter for Class 12 NCERT

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Atoms is the 12th chapter for Class 12 NCERT

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