SOLUTION TO NCERT PHYSICS (CLASS XII)

SOLUTION TO NCERT PHYSICS (CLASS XII)

Dr. Manjeet Singh M.Sc. (Physics), Ph.D., NET-JRF, GATE, Haryana SLET, Rajasthan SLET, H.P. SLET Qualified

2/25, Ansari Road, Darya Ganj, New Delhi-110 002

SOLUTION TO NCERT PHYSICS (CLASS XII) First Edition : 2015

ISBN: 978-93-83917-33-4

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Preface Physics is a mandatory subject for all XII (Non-Medical & Medical) students, where almost all the important elements of the subject are covered. The book entitled: Solution to NCERT Physics has been written as a help book of physics for XII class students. This help book contain fifteen chapters: Electric charges and fields, Electrostatic potential and capacitance, Current electricity, Moving charges and magnetism, Magnetism and matter, Electromagnetic induction, Alternating current, Electromagnetic waves, Ray optics and optical instruments, Wave optics, Dual nature of radiation and matter, Atoms, Nuclei, Semiconductor electronics: Materials, Devices and circuits, Communication systems. Following are the salient features of the book: • It represents a simple language and lucid style the fundamental principles of Physics in a manner such that the student may feel no difficulty in following them. • Each question is self contained and solved in a comprehensive style. • The mathematical calculations are clear and complete without any jumping and have been made simple, logical and easily understandable. • S.I. unit have been used throughout the text. However, C.G.S. units have been used, wherever necessary. • The point of view of students has always been kept in mind while preparing the present solution and hence it is hoped that the reader will experience no difficulty, whatsoever, in understanding the subject. I am greatly indebted to Prof. Praveen Aghamkar, Department of Physics, Chaudhary DeviLal University, Sirsa for his interest, inspiring guidance, and encouragement in this endeavor. I am thankful to Dr. Surender Duhan, Assistant Prof., Department of Material Science & Nano Technology, Deenbandhu Chhotu Ram University of Science & Technology, Murthal (Sonepat), Dr. Navneet Singh, Assistant Prof., Department of Physics, A.I.J.H.M. College, Rohtak, Mr. Jaivir Suhag, Mr. Vipin Pal Singh, J.V.M.G.R.R. College, Charkhi Dadri, for the help during the time when this book was written. I am very much indebted to many friends, Mr. Anil Bansal, Scientist C, IRDE (DRDO), Dehradun, Mr. Leeladhar, Scientist C, LASTEC (DRDO), New Delhi, Dr. Raj Kumar, Scientist C, CSIO (CSIR), Chandigarh, Mr. Devender Dudy, Government College, Birohar for their encouragement. I am thankful to Prof. Devendra Mohan, Prof. Ashish Aggarwal, Prof. Sujata Sanghi, Prof. Sneh Lata Goyal, Dr. David Joseph, Dr. Rajesh Punia and Dr. Neetu Ahlawat, Department of Applied Physics, Guru Jambheshwar University of Science & Technology, Hisar for their inspiration. I am very thankful to Mr. Ajay Chahal, Research Scholar, Indian Institute of Technology, Chennai, Mr. Kuldeep, Research Scholar, Indian Institute of Technology, Chennai and Mr. Tarun Garg, Research Scholar, Indian Institute of Technology, Bombay for the help whenever needed. Thanks to R.K. Jain, my publisher who encouraged me. Author

CONTENTS Preface ............................................................................................................................................. v

CHAPTER 1: (ELECTRIC CHARGES AND FIELDS).........................................................................................(1 - 19) CHAPTER 2: (ELECTROSTATIC POTENTIAL AND CAPACITANCE)........................................................(20 - 42) CHAPTER 3: (CURRENT ELECTRICITY).........................................................................................................(43 - 59) CHAPTER 4 : (MOVING CHARGES AND MAGNETISM).............................................................................(60 - 75) CHAPTER 5 : (MAGNETISM AND MATTER)..................................................................................................(76 - 88) CHAPTER 6: (ELECTROMAGNETIC INDUCTION)....................................................................................(89 - 100) CHAPTER 7 : (ALTERNATING CURRENT)...................................................................................................(101 - 117) CHAPTER 8: (ELECTROMAGNETIC WAVES).............................................................................................(118 - 126) CHAPTER 9: (RAY OPTICS AND OPTICAL INSTRUMENTS).................................................................(127 - 151) CHAPTER 10: (WAVE OPTICS)......................................................................................................................(152 - 161) CHAPTER 11: (DUAL NATURE OF RADIATION AND MATTER)..............................................................(162 - 182) CHAPTER 12: (ATOMS)..................................................................................................................................(183 - 193) CHAPTER 13: (NUCLEI)..................................................................................................................................(194 - 214) CHAPTER 14: (SEMICONDUCTOR ELECTRONICS: MATERIALS, DEVICES & CIRCUITS)..........(215 - 225) CHAPTER 15: (COMMUNICATION SYSTEMS)........................................................................................(226 - 230)

Electric Charges and Fields

1

C H A P TE R

1

ELECTRIC CHARGES AND FIELDS CHAPTER AT A GLANCE

1. Electric and magnetic forces determine the properties of atoms, molecules and bulk matter. 2. From simple experiments on frictional electricity, one can infer that there are two types of charges in nature; and that like charges repel and unlike charges attract. By convention, the charge on a glass rod rubbed with silk is positive; that on a plastic rod rubbed with fur is then negative. 3. Conductors allow movement of electric charge through them, insulators do not. In metals, the mobile charges are electrons; in electrolytes both positive and negative ions are mobile. 4. Electric charge has three basic properties: quantisation, additivity and conservation. Quantisation of electric charge means that total charge (q) of a body is always an integral multiple of a basic quantum of charge (e), i.e. q = ne, where n = 0, ± 1, ± 2, ± 3, .... Proton and electron have charges +e, –e, respectively. For macroscopic charges for which n is a very large number, quantisation of charge can be ignored. Additivity of electric charges means that the total charge of a system is the algebraic sum (i.e., the sum taking into account proper signs) of all individual charges in the system. Conservation of electric charges means that the total charge of an isolated system remains unchanged with time. This means that when bodies are charged through friction, there is a transfer of electric charge from one body to another, but no creation or destruction of charge. 5. Coulomb’s Law: The mutual electrostatic force between two point charges q1 and q2 is proportional to the product q1q2 and inversely proportional to the square of the distance r21 separating them. F21 = force on q2 due to q1 =

Mathematically,

k ( q1 q2 ) rˆ21 r212

where rˆ21 is a unit vector in the direction from q1 to q2 and k

1 4

is the constant of proportionality.. 0

In SI units, the unit of charge is coulomb. The experimental value of the constant 0

0

is:

= 8.854 × 10–12 C2N–1m–2

The approximate value of k is: k = 9 × 109 Nm2C–2 6. The ratio of electric force and gravitational force between a proton and an electron is: ke2 Gme m p

2.4 1039

2

Solution to NCERT Physics: Class XII

7. Superposition Principle: The principle is based on the property that the forces with which two charges attract or repel each other are not affected by the presence of a third (or more) additional charge(s). For an assembly of charges q1, q2, q3, ..., the force on any charge, say q1, is the vector sum of the force on q1 due to q2, the force on q1 due to q3, and so on. For each pair, the force is given by the Coulomb’s law for two charges stated earlier. 8. The electric field E at a point due to a charge configuration is the force on a small positive test charge q placed at the point divided by the magnitude of the charge. Electric field due to a point charge q has a magnitude q / 4

0

r 2 ; it is radially outwards from q, if q is positive, and radially inwards if q is

negative. Like Coulomb force, electric field also satisfies superposition principle. 9. An electric field line is a curve drawn in such a way that the tangent at each point on the curve gives the direction of electric field at that point. The relative closeness of field lines indicates the relative strength of electric field at different points; they crowd near each other in regions of strong electric field and are far apart where the electric field is weak. In regions of constant electric field, the field lines are uniformly spaced parallel straight lines. 10. Some of the important properties of field lines are: (i) Field lines are continuous curves without any breaks. (ii) Two field lines cannot cross each other. (iii) Electrostatic field lines start at positive charges and end at negative charges —they cannot form closed loops. 11. An electric dipole is a pair of equal and opposite charges q and –q separated by some distance 2a. Its dipole moment vector p has magnitude 2qa and is in the direction of the dipole axis from –q to q. 12. Field of an electric dipole in its equatorial plane (i.e., the plane perpendicular to its axis and passing through its centre) at a distance r from the centre: E

p (a2

4

0

4

p 3 0r

1 r 2 )3 / 2

for r

a

Dipole electric field on the axis at a distance r from the centre: E

2 pr 1 2 4 0 (r a 2 )2 2p 4 0 r3

for r

a

The 1/r3 dependence of dipole electric fields should be noted in contrast to the 1/r2 dependence of electric field due to a point charge. 13. In a uniform electric field E , a dipole experiences a torque

given by

p E

but experiences no net force. 14. The flux

of electric field E through a small area element S is given by E. S

Electric Charges and Fields

3

The vector area element S is: S (nˆ)

S

where S is the magnitude of the area element and nˆ is normal to the area element, which can be considered planar for sufficiently small S . For an area element of a closed surface, nˆ is taken to be the direction of outward normal, by convention. 15. Gauss’s law: The flux of electric field through any closed surface S is 1/

0

times the total charge

enclosed by S. The law is especially useful in determining electric field E , when the source distribution has simple symmetry: (i) Thin infinitely long straight wire of uniform linear charge density E

nˆ

2

0

where r is the perpendicular distance of the point from the wire and nˆ is the radial unit vector in the plane normal to the wire passing through the point. (ii) Infinite thin plane sheet of uniform surface charge density . E

2

nˆ 0

where nˆ is a unit vector normal to the plane, outward on either side. (iii) Thin spherical shell of uniform surface charge density . E

q 4

0

r2

rˆ ( r

R)

E 0 ( r R) where r is the distance of the point from the centre of the shell and R the radius of the shell. q is the total

charge of the shell: q 4 r 2 . The electric field outside the shell is as though the total charge is concentrated at the centre. The same result is true for a solid sphere of uniform volume charge density. The field is zero at all points inside the shell.

EXERCISE Q. 1.1: What is the force between two small charges spheres having charges of 2 × 10–7 C and 3 × 10–7 C placed 30 cm apart in air? Answer: Charge on the first sphere, q1 = 2 × 10–7 C Charge on the second sphere, q2 = 3 × 10–7 C Distance between the spheres, r = 30 cm = 0.3 m According to Coulomb’s law, electrostatic force between the spheres is given by F

q1 q2 4 0r 2

9 109

2 10 7 (0.3)2

3 10

7

6 10 3 N

4

Solution to NCERT Physics: Class XII

Hence, force between the two small charged spheres is 6×10–3 N. The charges are of same nature. Hence, force between them will be repulsive. Q. 1.2: The electrostatic force on a small sphere of charge 0.4 C due to another small sphere of charge –0.8 C in air is 0.2 N. (a) What is the distance between the two spheres? (b) What is the force on the second sphere due to the first? Answer: (a) Electrostatic force on the first sphere, F = 0.2 N Charge on this sphere, q1 = 0.4 C = 0.4 × 10–6 C Charge on the second sphere, q1 = –0.8

C = –0.8 × 10–6 C

According to Coulomb’s law, electrostatic force between the spheres is given by: F

r2

q1q2 4 0F

0.4 10

6

8 10 0.2

6

9 109

144 10

4

r

144 10

4

q1q2 4 0r2 0.12 m.

The distance between the two spheres is 0.12 m. (b) Both the spheres attract each other with the same force. Therefore, the force on the second sphere due to the first is 0.2 N. Q. 1.3: Check that the ratio ke2/G memp is dimensionless. Look up a Table of Physical Constants and determine the value of this ratio. What does the ratio signify? Answer: Given ratio is:

ke 2 Gme m p

where, G = Gravitational constant. Its unit is Nm2kg–2. me and mp = Masses of electron and proton. Their unit is kg. e = Electric charge. Its unit is C. k = A constant = 0

1 4

0

= Permittivity of free space. Its unit is Nm2C–2.

Unit of the given ratio is:

ke2 Gme m p

[Nm2 C 2 ][C2 ] [Nm 2 kg 2 ][kg][kg]

M 0 L0 T0

Hence, the given ratio is dimensionless. e = 1.6×10–19 C, G = 6.67×10–11 Nm2kg-2, me = 9.1×10–31 kg, mp = 1.66×10–27 kg. ke2 Gme m p

9 109 (1.6 10 19 ) 2 6.67 10 11 9.1 10 3 1.67 10

22

2.3 1039 .

This is the ratio of electric force to the graviational force between a proton and an electron, keeping distance between them constant.

Electric Charges and Fields

7

Q. 1.10: An electric dipole with dipole moment 4 × 10–9 Cm is aligned at 30° with the direction of a uniform electric field of magnitude 5 × 104 NC–1. Calculate the magnitude of the torque acting on the dipole. Answer: Electric dipole moment, p = 4 × 10–9 Cm Angle made by p with a uniform electric field, = 30° Electric field intensity, E = 5 × 104 NC–1 Torque acting on the dipole,

pE sin

4 10

9

5 104 sin

20 10

5

1 10 4 Nm. 2

Therefore, the magnitude of the torque acting on the dipole is 10–4 Nm. Q. 1.11: A polythene piece rubbed with wool is found to have a negative charge of 3 × 10–7 C. (a) Estimate the number of electrons transferred (from which to which?) (b) Is there a transfer of mass from wool to polythene? Answer: (a) When polythene is rubbed against wool, a number of electrons get transferred from wool to polythene. Hence, wool becomes positively charged and polythene becomes negatively charged. Amount of charge on the polythene place, q = – 3 × 10–7 C Amount of charge on an electron, e = –1.6 × 10–19 C Let, number of electrons transferred from wool to polythene = n n

According to quantisation of charge, q = ne

q e

3 10 7 1.6 10 19

1.87 1012 .

Therefore, the number of electrons transferred from wool to polythene is 1.87 × 1012. (b) Yes, there is a transfer of mass taking place. This is because an electron has mass, me = 9.1 × 10–3 kg. Total mass transferred to polythene from wool, m = me × n = 9.1 × 10–31 × 1.85 × 1012 = 1.706 × 10–18 kg. Hence, a negligible amount of mass is transferred from wool to polythene. Q. 1.12: (a) Two insulated charged copper spheres A and B have their centers separated by a distance of 50 cm. What is the mutual force of electrostatic repulsion if the charge on each is 6.5 × 10–7 C? The radii of A and B are negligible compared to the distance of separation. (b) What is the force of repulsion if each sphere is charted double the above amount, and the distance between them is halved? Answer:(a) Charge on sphere A, qA = Charge on sphere B, qB = 6.5 × 10–7 C. Distance between the spheres, r = 50 cm = 0.5 m 9 109 (6.5 10 7 ) 2 qA qB = 1.52 × 10–2 N. (0.5) 2 4 0r2 Therefore, the force between the two spheres is 1.52 × 10–2 N. (b) After doubling charge, Charge on sphere B, qB = 2 × 6.5 × 10–7 C = 1.3 × 10–6 C.

Force of repulsion between the two spheres, F

The distance between the spheres is halved, i.e., r = Force of repulsion between two spheres, F

qA qB 4 0r2

0.5 2

0.25 m.

9 109 1.3 10 6 1.3 10 (0.25) 2

Therefore, the force between the two spheres is 0.243 N.

6

= 0.243 N.

8

Solution to NCERT Physics: Class XII

Q. 1.13: Suppose the spheres A and B in Q. 1.12 have identical sizes. A third sphere of the same size but uncharged is brought in contact with the first, then brought in contact with the second, and finally removed from both. What is the new force of repulsion between A and B? Answer: Distance between the spheres, A and B, r = 0.5 m Initially, the charge on each sphere, q = 6.5 × 10–7 C. When sphere A is touched with an uncharged sphere C, C. Hence, charge on each of the spheres, A and C, is When sphere C with charge

q . 2

q is brought in contact with sphere B with charge q, total charges on the 2

q system will divide into two equal haves given as: 2

q 2

Force of repulsion between sphere A having charge q 3q 2 4 4 0r2

3q 2 8 4 0r2

q amount of charge from a will transfer to sphere 2

9 109

3q . 4

q 3q and sphere B having charge 2 4

3 (6.5 10 7 ) 2 = 5.703 × 10–3 N 8 (0.5) 2

Therefore, the force of attraction between the two spheres is 5.703 × 10–3 N. Q. 1.14: Shows tracks of three charged particles in a uniform electrostatic feld. Give the signs of the three charges. Which particle has the highest charge to mass ratio? – – + + + + + + + + +

1 2

y x

– – – – – – – – –

3

Answer: Opposite charges attract each other and same charges repel each other. It can be observed that particles 1 and 2 both move towards the positively charged plate and repel away from the negatively charged plate. Hence, these two particles are negatively charged. It can also be observed that particle 3 moves towards the negatively charged plate and repels away from the positively charged plate. Hence, particle 3 is positively charged. The charge to mass ratio (emf) is directly proportional to the displacement or amount of deflection for a given velocity. Since the deflection of particle 3 is the maximum, it has the highest charge to mass ratio.

Electric Charges and Fields

9

Q. 1.15: Consider a uniform electric field E = 3 ×103 iˆ NC-1. (a) What is the flux of this field through a square of 10 cm on a side whose plane is parallel to the yz plane? (b) What is the flux through the same square if the normal to its place makes a 60° angle with x-axis? Answer: (a) Electric field intensity, E 3 103 iˆ NC-1 Magnitude of electric field intensity, E

3 103 NC-1

Side of the square, s = 10 cm = 0.1 m Area of the square, A = s2 = 0.01 m2 The plane of the square is parallel to the y-z plane. Hence, angle between the unit vector normal to the plane and electric field, = 0°. Flux ( ) through the plane is:

| E | A cos

= 3 × 103 × 0.01 × cos 0° = 30 Nm2C-1

(b) Plane makes an angle of 60° with the x-axis. Hence,

= 60°

1 15 Nm2C-1 2 Q. 1.16: What is the net flux of the uniform electric field of Q. 1.15 through a cube of side 20 cm oriented so that its faces are parallel to the coordinate planes? Answer: All the faces of a cube are parallel to the coordinate axes. Therefore, the number of field lines entering the cube is equal to the number of field lines piercing out of the cube. As a result, net flux through the cube is zero. Q. 1.17: Careful measurement of the electric field at the surface of a black box indicates that the net outward flux through the surface of the box is 8.0×103 Nm2C-1. (a) What is the net charge inside the box? (b) If the net outward flux through the surface of the box were zero, could you conclude that there were no charges inside the box? Why or why not?

Flux,

| E | A cos

= 3 × 103 × 0.01 × cos 60° = 30

Answer: Net outward flux through the surface of the box, For a body containing net charge q, electric flux,

= 8.0 × 103 Nm2C-1

q 0

q

0

= 8.854 × 10–12 × 8.0 × 103 = 7.08 × 10–8 = 0.07 C 5 cm

Q. 1.18: A point charge + 10 C is a distance 5 cm directly above the centre of a square of side 10 cm, as shown in Fig. What is the magnitude of the electric flux through the square? (Hint: Think of the square as one face of a cube with edge 10 cm.) Answer: The square can be considered as one face of a cube of edge 10 cm with a centre where charge q is placed.

10 c m

Therefore, the net charge inside the box is 0.07 C . (b) No. Net flux piercing out through a body depends on the net charge contained in the body. If net flux is zero, then it can be inferred that net charge inside the body is zero. The body may have equal amount of positive and negative charges.

10 cm