Physics 12th Book 1,2

Physics Example 14.7 Why are Si and GaAs are preferred materials for solar cells? Solution The solar radiation spectrum received by us is shown in Fig. 14.26. FIGURE 14.26 The maxima is near 1.5 eV. For photo-excitation, hν > Eggiv. eHbeentcteer, semiconductor with band gap ~1.5 eV or lower is likely to solar conversion efficiency. Silicon has Eg ~ 1.1 eV while for GaAs it is ~1.53 eV. In fact, GaAs is better (in spite of its higher band gap) than Si because of its relatively higher absorption coefficient. If we choose materials like CdS or CthdeSeso(Elagr ~ 2.4 eV), we can use only the high energy component of energy for photo-conversion and a significant part of energy will be of no use. EXAMPLE 14.7 The question arises: why we do not use material like PbS (Eg ~ 0.4 eV) which satisfy the condition hν > Esog ,fomr oνstmoafxtihmeascoolarrrersapdoiantdioinngwtoillthbee solar radiation spectra? If we do absorbed on the top-layer of solar cell and will not reach in or near the depletion region. For effective electron-hole separation, due to the junction field, we want the photo-generation to occur in the junction region only. 490 14.9 DIGITAL ELECTRONICS AND LOGIC GATES In electronics circuits like amplifiers, oscillators, introduced to you in earlier sections, the signal (current or voltage) has been in the form of continuous, time-varying voltage or current. Such signals are called continuous or analog signals. A typical analog signal is shown in Figure. 14.27(a). Fig. 14.27(b) shows a pulse waveform in which only discrete values of voltages are possible. It is convenient to use binary numbers to represent such signals. A binary number has only two digits ‘0’ (say, 0V) and ‘1’ (say, 5V). In digital electronics we use only these two levels of voltage as shown in Fig. 14.27(b). Such signals are called Digital Signals. 2020-21

Semiconductor Electronics: Materials, Devices and Simple Circuits In digital circuits only two values (represented by 0 or 1) of the input and output voltage are permissible. This section is intended to provide the first step in our understanding of digital electronics. We shall restrict our study to some basic building blocks of digital electronics (called Logic Gates) which process the digital signals in a specific manner. Logic gates are used in calculators, digital watches, computers, robots, industrial control systems, and in telecommunications. A light switch in your house can be used as an example of a digital circuit. The light is either ON or OFF depending on the switch position. When the light is ON, the output value is ‘1’. When the light is OFF the output value is ‘0’. The inputs are the position of the light switch. The switch is placed either in the ON or OFF position to activate the light. FIGURE 14.27 (a) Analog signal, (b) Digital signal. 14.9.1 Logic gates Input Output A gate is a digital circuit that follows curtain logical relationship AY between the input and output voltages. Therefore, they are generally known as logic gates — gates because they control the flow of 01 information. The five common logic gates used are NOT, AND, OR, NAND, NOR. Each logic gate is indicated by a symbol and its function 10 is defined by a truth table that shows all the possible input logic level combinations with their respective output logic levels. Truth tables (b) help understand the behaviour of logic gates. These logic gates can be realised using semiconductor devices. FIGURE 14.28 (a) Logic symbol, (i) NOT gate (b) Truth table of This is the most basic gate, with one input and one output. It produces NOT gate. a ‘1’ output if the input is ‘0’ and vice-versa. That is, it produces an inverted version of the input at its output. This is why it is also known 491 as an inverter. The commonly used symbol together with the truth table for this gate is given in Fig. 14.28. (ii) OR Gate An OR gate has two or more inputs with one output. The logic symbol and truth table are shown in Fig. 14.29. The output Y is 1 when either input A or input B or both are 1s, that is, if any of the input is high, the output is high. 2020-21

Physics Input Output AB Y 00 0 01 1 10 1 11 1 (b) FIGURE 14.29 (a) Logic symbol (b) Truth table of OR gate. Apart from carrying out the above mathematical logic operation, this gate can be used for modifying the pulse waveform as explained in the following example. Example 14.8 Justify the output waveform (Y) of the OR gate for the following inputs A and B given in Fig. 14.30. Solution Note the following: • At t < t1; A = 0, B = 0; Hence Y = 0 • For t1 to t2; A = 1, B = 0; Hence Y = 1 • For t2 to t3; A = 1, B = 1; Hence Y = 1 • For t3 to t4; A = 0, B = 1; Hence Y = 1 • For t4 to t5; A = 0, B = 0; Hence Y = 0 • For t5 to t6; A = 1, B = 0; Hence Y = 1 • For t > t6; A = 0, B = 1; Hence Y = 1 Therefore the waveform Y will be as shown in the Fig. 14.30. EXAMPLE 14.8 FIGURE 14.30 Input Output (iii) AND Gate An AND gate has two or more inputs and one output. The output Y of AB Y AND gate is 1 only when input A and input B are both 1. The logic 00 0 symbol and truth table for this gate are given in Fig. 14.31 01 0 10 0 FIGURE 14.31 (a) Logic symbol, (b) Truth table of AND gate. 11 1 (b) 492 2020-21

Semiconductor Electronics: Materials, Devices and Simple Circuits Example 14.9 Take A and B input waveforms similar to that in Example 14.8. Sketch the output waveform obtained from AND gate. Solution • For t ≤t1; A = 0, B = 0; Hence Y = 0 • For t1 to t2; A = 1, B = 0; Hence Y = 0 • For t2 to t3; A = 1, B = 1; Hence Y = 1 • For t3 to t4; A = 0, B = 1; Hence Y = 0 • For t4 to t5; A = 0, B = 0; Hence Y = 0 • For t5 to t6; A = 1, B = 0; Hence Y = 0 • For t > t6; A = 0, B = 1; Hence Y = 0 Based on the above, the output waveform for AND gate can be drawn as given below. FIGURE 14.32 EXAMPLE 14.9 (iv) NAND Gate This is an AND gate followed by a NOT gate. If inputs A and B are both ‘1’, the output Y is not ‘1’. The gate gets its name from this NOT AND behaviour. Figure 14.33 shows the symbol and truth table of NAND gate. NAND gates are also called Universal Gates since by using these gates you can realise other basic gates like OR, AND and NOT (Exercises 14.12 and 14.13). Input Output AB Y 00 1 01 1 10 1 11 0 (b) FIGURE 14.33 (a) Logic symbol, (b) Truth table of NAND gate. Example 14.10 Sketch the output Y from a NAND gate having inputs EXAMPLE 14.10 A and B given below: Solution A = 1, B = 1; Hence Y = 0 493 • For t < t1; • For t1 to t2; A = 0, B = 0; Hence Y = 1 • For t2 to t3; A = 0, B = 1; Hence Y = 1 • For t3 to t4; A = 1, B = 0; Hence Y = 1 2020-21

Physics A = 1, B = 1; Hence Y = 0 A = 0, B = 0; Hence Y = 1 • For t4 to t5; A = 0, B = 1; Hence Y = 1 • For t5 to t6; • For t > t6; EXAMPLE 14.10 FIGURE 14.34 (v) NOR Gate It has two or more inputs and one output. A NOT- operation applied after OR gate gives a NOT-OR gate (or simply NOR gate). Its output Y is ‘1’ only when both inputs A and B are ‘0’, i.e., neither one input nor the other is ‘1’. The symbol and truth table for NOR gate is given in Fig. 14.35. Input Output AB Y 00 1 01 0 10 0 11 0 (b) FIGURE 14.35 (a) Logic symbol, (b) Truth table of NOR gate. NOR gates are considered as universal gates because you can obtain all the gates like AND, OR, NOT by using only NOR gates (Exercises 14.14 and 14.15). FASTER AND SMALLER: THE FUTURE OF COMPUTER TECHNOLOGY The Integrated Chip (IC) is at the heart of all computer systems. In fact ICs are found in almost all electrical devices like cars, televisions, CD players, cell phones etc. The miniaturisation that made the modern personal computer possible could never have happened without the IC. ICs are electronic devices that contain many transistors, resistors, capacitors, connecting wires – all in one package. You must have heard of the 494 2020-21

Semiconductor Electronics: Materials, Devices and Simple Circuits microprocessor. The microprocessor is an IC that processes all information in a computer, like keeping track of what keys are pressed, running programmes, games etc. The IC was first invented by Jack Kilky at Texas Instruments in 1958 and he was awarded Nobel Prize for this in 2000. ICs are produced on a piece of semiconductor crystal (or chip) by a process called photolithography. Thus, the entire Information Technology (IT) industry hinges on semiconductors. Over the years, the complexity of ICs has increased while the size of its features continued to shrink. In the past five decades, a dramatic miniaturisation in computer technology has made modern day computers faster and smaller. In the 1970s, Gordon Moore, co-founder of INTEL, pointed out that the memory capacity of a chip (IC) approximately doubled every one and a half years. This is popularly known as Moore’s law. The number of transistors per chip has risen exponentially and each year computers are becoming more powerful, yet cheaper than the year before. It is intimated from current trends that the computers available in 2020 will operate at 40 GHz (40,000 MHz) and would be much smaller, more efficient and less expensive than present day computers. The explosive growth in the semiconductor industry and computer technology is best expressed by a famous quote from Gordon Moore: “If the auto industry advanced as rapidly as the semiconductor industry, a Rolls Royce would get half a million miles per gallon, and it would be cheaper to throw it away than to park it”. SUMMARY 1. Semiconductors are the basic materials used in the present solid state electronic devices like diode, transistor, ICs, etc. 2. Lattice structure and the atomic structure of constituent elements decide whether a particular material will be insulator, metal or semiconductor. 3. Metals have low resistivity (10–2 to 10–8 Ω m), insulators have very high resistivity (>108 Ω m–1), while semiconductors have intermediate values of resistivity. 4. Semiconductors are elemental (Si, Ge) as well as compound (GaAs, CdS, etc.). 5. Pure semiconductors are called ‘intrinsic semiconductors’. The presence of charge carriers (electrons and holes) is an ‘intrinsic’ property of the material and these are obtained as a result of thermal excitation. The ncounmdbuecrtoorfse.leHcotrleosnsar(neee) sisseenqtuiaallltyoetlheectnrounmvbaecraonf chioelsesw(inthh ) in intrinsic an effective positive charge. 6. The number of charge carriers can be changed by ‘doping’ of a suitable impurity in pure semiconductors. Such semiconductors are known as extrinsic semiconductors. These are of two types (n-type and p-type). 7. In n-type semiconductors, ne >> nh while in p-type semiconductors nh >> ne. 8. n-type semiconducting Si or Ge is obtained by doping with pentavalent atoms (donors) like As, Sb, P, etc., while p-type Si or Ge can be obtained by doping with trivalent atom (acceptors) like B, Al, In etc. 9. nneenuhtr=alnitiy2 .in all cases. Further, the material possesses an overall charge 495 2020-21

Physics 10. There are two distinct band of energies (called valence band and conduction band) in which the electrons in a material lie. Valence band energies are low as compared to conduction band energies. All energy levels in the valence band are filled while energy levels in the conduction band may be fully empty or partially filled. The electrons in the conduction band are free to move in a solid and are responsible for the conductivity. The extent of conductivity depends upon the energy cghoaenpatd,(uElicggt)hiobtneotrbweaelneecdntrEtihCc.aeTl tehonepeerolgefycvttraooltenhnsecfecroobnmadnvudaclte(iEnonVce)babananndddathcnaednbthbouettseo,xmpcritooedfduthbcyee a change in the current flowing in a semiconductor. 11. For insulators Eg > 3 eV, for semiconductors Eg is 0.2 eV to 3 eV, while for metals Eg ≈ 0. 12. p-n junction is the ‘key’ to all semiconductor devices. When such a junction is made, a ‘depletion layer’ is formed consisting of immobile ion-cores devoid of their electrons or holes. This is responsible for a junction potential barrier. 13. By changing the external applied voltage, junction barriers can be changed. In forward bias (n-side is connected to negative terminal of the battery and p-side is connected to the positive), the barrier is decreased while the barrier increases in reverse bias. Hence, forward bias current is more (mA) while it is very small (µA) in a p-n junction diode. 14. Diodes can be used for rectifying an ac voltage (restricting the ac voltage to one direction). With the help of a capacitor or a suitable filter, a dc voltage can be obtained. 15. There are some special purpose diodes. 16. Zener diode is one such special purpose diode. In reverse bias, after a certain voltage, the current suddenly increases (breakdown voltage) in a Zener diode. This property has been used to obtain voltage regulation. 17. p-n junctions have also been used to obtain many photonic or optoelectronic devices where one of the participating entity is ‘photon’: (a) Photodiodes in which photon excitation results in a change of reverse saturation current which helps us to measure light intensity; (b) Solar cells which convert photon energy into electricity; (c) Light Emitting Diode and Diode Laser in which electron excitation by a bias voltage results in the generation of light. 18. There are some special circuits which handle the digital data consisting of 0 and 1 levels. This forms the subject of Digital Electronics. 19. The important digital circuits performing special logic operations are called logic gates. These are: OR, AND, NOT, NAND, and NOR gates. 496 POINTS TO PONDER 1. wThheicehnemrgeaynbsanthdast(EthCeosreEaVr)einnotht elosceamteidcoinndauncytosrpseacrieficsppalacceedienloscidaelisthede solid. The energies are the overall averages. When you see a picture in rwehspicehctiEvCelyortaEkeVnasriemdplryaawsnthaesbsotttroamigohftcloinndeus,cttihonenbatnhdeyensehrgoyulledveblse and top of valence band energy levels. 2. In elemental semiconductors (Si or Ge), the n-type or p-type semiconductors are obtained by introducing ‘dopants’ as defects. In compound semiconductors, the change in relative stoichiometric ratio can also change the type of semiconductor. For example, in ideal GaAs 2020-21

Semiconductor Electronics: Materials, Devices and Simple Circuits the ratio of Ga:As is 1:1 but in Ga-rich or As-rich GaAs it could respectively be Ga1.1 As0.9 or Ga0.9 As1.1. In general, the presence of defects control the properties of semiconductors in many ways. 3. In modern day circuit, many logical gates or circuits are integrated in one single ‘Chip’. These are known as Intgrated circuits (IC). EXERCISES 14.1 In an n-type silicon, which of the following statement is true: 14.2 (a) Electrons are majority carriers and trivalent atoms are the 14.3 dopants. 14.4 (b) Electrons are minority carriers and pentavalent atoms are the 14.5 dopants. 14.6 14.7 (c) Holes are minority carriers and pentavalent atoms are the dopants. (d) Holes are majority carriers and trivalent atoms are the dopants. Which of the statements given in Exercise 14.1 is true for p-type semiconductos. Carbon, silicon and germanium have four valence electrons each. These are characterised by valence and conduction bands separated by energy band gap respectively equal to (Eg)C, (Eg)Si and (Eg)Ge. Which of the following statements is true? (a) (Eg)Si < (Eg)Ge < (Eg)C (b) (Eg)C < (Eg)Ge > (Eg)Si (c) (Eg)C > (Eg)Si > (Eg)Ge (d) (Eg)C = (Eg)Si = (Eg)Ge In an unbiased p-n junction, holes diffuse from the p-region to n-region because (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. 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. 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. A p-n photodiode is fabricated from a semiconductor with band gap of 2.8 eV. Can it detect a wavelength of 6000 nm? 497 2020-21

Physics ADDITIONAL EXERCISES 14.8 The number of silicon atoms per m3 is 5 × 1028. This is doped 14.9 simultaneously with 5 × 1022 atoms per m3 of Arsenic and 5 × 1020 per m3 atoms of Indium. Calculate the number of electrons and holes. Given that ni = 1.5 × 1016 m–3. Is the material n-type or p-type? In an intrinsic semiconductor ethleecternonergmyobgailpityEag nids 1.2eV. Its hole mobility is much smaller than independent of temperature. What is the ratio between conductivity at 600K and that at 300K? Assume that the temperature dependence of intrinsic carrier concentration ni is given by where n0 is a constant. 14.10 In a p-n junction diode, the current I can be expressed as where It0hies called the reverse saturation current, V is the voltage across diode and is positive for forward bias and negative for reverse bias, and I is the current through the diode, akBb is l the Boltzmann constant (8.6×10–5 eV/K) and T is the so ute temperature. If for a given diode I0 = 5 × 10–12 A and T = 300 K, then (a) What will be the forward current at a forward voltage of 0.6 V? (b) What will be the increase in the current if the voltage across the diode is increased to 0.7 V? (c) What is the dynamic resistance? (d) What will be the current if reverse bias voltage changes from 1 V to 2 V? 14.11 You are given the two circuits as shown in Fig. 14.36. Show that circuit (a) acts as OR gate while the circuit (b) acts as AND gate. 498 FIGURE 14.36 14.12 Write the truth table for a NAND gate connected as given in Fig. 14.37. FIGURE 14.37 Hence identify the exact logic operation carried out by this circuit. 2020-21

Semiconductor Electronics: Materials, Devices and Simple Circuits 14.13 You are given two circuits as shown in Fig. 14.38, which consist of NAND gates. Identify the logic operation carried out by the two circuits. FIGURE 14.38 14.14 Write the truth table for circuit given in Fig. 14.39 below consisting of NOR gates and identify the logic operation (OR, AND, NOT) which this circuit is performing. FIGURE 14.39 (Hint: A = 0, B = 1 then A and B inputs of second NOR gate will be 0 and hence Y=1. Similarly work out the values of Y for other combinations of A and B. Compare with the truth table of OR, AND, NOT gates and find the correct one.) 14.15 Write the truth table for the circuits given in Fig. 14.40 consisting of NOR gates only. Identify the logic operations (OR, AND, NOT) performed by the two circuits. FIGURE 14.40 499 2020-21

Physics ANSWERS CHAPTER 1 1.1 6 × 10–3 N (repulsive) 1.2 (a) 12 cm 1.3 (b) 0.2 N (attractive) 1.5 2.4 × 1039. This is the ratio of electric force to the gravitational force (at the same distance) between an electron and a proton. 1.6 1.8 Charge is not created or destroyed. It is merely transferred from one body to another. 1.9 1.10 Zero N 1.11 (a) 5.4 × 106 N C–1 along OB 1.12 (b) 8.1 × 10–3 N along OA 1.13 Total charge is zero. Dipole moment = 7.5 × 10–8 C m along z-axis. 1.14 10–4 N m 1.15 1.16 (a) 2 × 1012, from wool to polythene. (b) Yes, but of a negligible amount ( = 2 × 10–18 kg in the example). 1.17 (a) 1.5 × 10–2 N 1.18 (b) 0.24 N 1.19 1.20 5.7 × 10–3 N 1.21 Charges 1 and 2 are negative, charge 3 is positive. Particle 3 has 1.22 the highest charge to mass ratio. 1.23 (a) 30Nm2/C, (b) 15 Nm2/C 1.24 Zero. The number of lines entering the cube is the same as the number of lines leaving the cube. (a) 0.07 µC (b) No, only that the net charge inside is zero. 2.2 × 105 N m2/C 1.9 × 105 N m2/C because the charge enclosed is the same in the (a) –103 N m2/C; two cases. (b) –8.8 nC – 6.67 nC (a) 1.45 × 10–3 C (b) 1.6 × 108 Nm2/C 10 µC/m 288 (a) Zero, (b) Zero, (c) 1.9 N/C 2020-21

Answers 1.25 9.81 × 10–4 mm. 1.26 1.27 Only (c) is right; the rest cannot represent electrostatic field lines, 1.28 (a) is wrong because field lines must be normal to a conductor, (b) is 1.29 wrong because field lines cannot start from a negative charge, (d) is wrong because field lines cannot intersect each other, (e) is 1.31 wrong because electrostatic field lines cannot form closed loops. 1.32 The force is 10–2 N in the negative z-direction, that is, in the direction of decreasing electric field. You can check that this is also the 1.34 direction of decreasing potential energy of the dipole; torque is zero. (a) Hint: Choose a Gaussian surface lying wholly within the conductor and enclosing the cavity. (b) Gauss’s law on the same surface as in (a) shows that q must induce –q on the inner surface of the conductor. (c) Enclose the instrument fully by a metallic surface. Hint: Consider the conductor with the hole filled up. Then the field just outside is (σ/ε0) nˆ and is zero inside. View this field as a superposition of the field due to the filled up hole plus the field due to the rest of the charged conductor. Inside the conductor, these fields are equal and opposite. Outside they are equal both in magnitude and direction. Hence, the field due to the rest of the conductor is σ nˆ .  2 ε0  p;uud; n;udd. (a) Hint: Prove it by contradiction. Suppose the equilibrium is stable; then the test charge displaced slightly in any direction will experience a restoring force towards the null-point. That is, all field lines near the null point should be directed inwards towards the null-point. That is, there is a net inward flux of electric field through a closed surface around the null-point. But by Gauss’s law, the flux of electric field through a surface, not enclosing any charge, must be zero. Hence, the equilibrium cannot be stable. (b) The mid-point of the line joining the two charges is a null-point. Displace a test charge from the null-point slightly along the line. There is a restoring force. But displace it, say, normal to the line. You will see that the net force takes it away from the null-point. Remember, stability of equilibrium needs restoring force in all directions. 1.6 cm CHAPTER 2 289 2.1 10 cm, 40 cm away from the positive charge on the side of the negative charge. 2.2 2.7 × 106 V 2.3 (a) The plane normal to AB and passing through its mid-point has zero potential everywhere. (b) Normal to the plane in the direction AB. 2.4 (a) Zero 2020-21

Physics 2.5 (b) 105 N C–1 2.6 (c) 4.4 × 104 N C–1 2.7 2.8 96 pF 2.9 2.10 (a) 3 pF 2.11 (b) 40 V 2.12 2.13 (a) 9 pF 2.14 (b) 2 × 10–10 C, 3 × 10–10 C, 4 × 10–10 C 18 pF, 1.8 × 10–9 C 2.15 (a) V = 100 V, C = 108 pF, Q = 1.08 × 10–8 C (b) Q = 1.8 × 10–9 C, C = 108 pF, V = 16.6 V 2.17 1.5 × 10–8 J 2.18 6 × 10–6 J 2.19 1.2 J; the point R is irrelevant to the answer. 2.20 Potential = 4q/( 3 π ε0 b ); field is zero, as expected by symmetry. (a) 2.4 × 105 V; 4.0 × 105 Vm–1 from charge 2.5 µC to 1.5 µC. 2.21 (b) 2.0 × 105 V; 6.6 × 105 Vm–1 in the direction that makes an angle 290 of about 69° to the line joining charge 2.5 µC to 1.5 µC. (a) − q /(4 π r12 ), (Q + q) / (4 π r22 ) (b) By Gauss’s law, the net charge on the inner surface enclosing the cavity (not having any charge) must be zero. For a cavity of arbitrary shape, this is not enough to claim that the electric field inside must be zero. The cavity may have positive and negative charges with total charge zero. To dispose of this possibility, take a closed loop, part of which is inside the cavity along a field line and the rest inside the conductor. Since field inside the conductor is zero, this gives a net work done by the field in carrying a test charge over a closed loop. We know this is impossible for an electrostatic field. Hence, there are no field lines inside the cavity (i.e., no field), and no charge on the inner surface of the conductor, whatever be its shape. λ/(2 π ε0 r ), where r is the distance of the point from the common axis of the cylinders. The field is radial, perpendicular to the axis. (a) –27.2 eV (b) 13.6 eV (c) –13.6 eV, 13.6 eV. Note in the latter choice the total energy of the hydrogen atom is zero. –19.2 eV; the zero of potential energy is taken to be at infinity. The ratio of electric field of the first to the second is (b/a ). A flat portion may be equated to a spherical surface of large radius, and a pointed portion to one of small radius. (a) On the axis of the dipole, potential is (± 1/4 π ε0) p/(x2 – a2) where p =2qa is the magnitude of the dipole moment; the + sign when the point is closer to q and the – sign when it is closer to – q. Normal to the axis, at points (x, y, 0), potential is zero. (b) The dependence on r is 1/r 2 type. (c) Zero. No, because work done by electrostatic field between two points is independent of the path connecting the two points. 2020-21

Answers 2.22 For large r, quadrupole potential goes like 1/r 3, dipole potential goes 291 2.23 like 1/r 2, monopole potential goes like 1/r. 2.24 Eighteen 1 µF capacitors arranged in 6 parallel rows, each row 2.25 consisting of 3 capacitors in series. 2.26 1130 km2 2.27 2.28 Equivalent capacitance = (200/3) pF. Q1 = 10 –8 C, V1 = 100 V ; Q2 = Q3 = 10 –8 C 2.30 V2 = V3 = 50 V 2.31 Q4 = 2.55 × 10 –8 C, V4 = 200 V (a) 2.55 × 10 –6 J 2.32 (b) u = 0.113 J m –3, u = (½) ε0 E 2 2.33 2.67 × 10 –2 J 2.34 Hint: Suppose we increase the separation of the plates by ∆x. Work 2.35 done (by external agency) = F ∆x. This goes to increase the potential 2.36 energy of the capacitor by u a ∆x where u is energy density. Therefore, F = u a which is easily seen to be (1/2) QE, using u = (1/2) ε0 E 2. The physical origin of the factor 1/2 in the force formula lies in the fact that just outside the conductor, field is E, and inside it is zero. So, the average value E/2 contributes to the force. (a) 5.5 × 10–9 F (b) 4.5 × 102 V (c) 1.3 × 10–11 F (a) No, because charge distributions on the spheres will not be uniform. (b) No. (c) Not necessarily. (True only if the field line is a straight line.) The field line gives the direction of acceleration, not that of velocity, in general. (d) Zero, no matter what the shape of the complete orbit. (e) No, potential is continuous. (f ) A single conductor is a capacitor with one of the ‘plates’ at infinity. (g) A water molecule has permanent dipole moment. However, detailed explanation of the value of dielectric constant requires microscopic theory and is beyond the scope of the book. 1.2 × 10–10 F, 2.9 × 104 V 19 cm2 (a) Planes parallel to x-y plane. (b) Same as in (a), except that planes differing by a fixed potential get closer as field increases. (c) Concentric spheres centred at the origin. (d) A periodically varying shape near the grid which gradually reaches the shape of planes parallel to the grid at far distances. Hint: By Gauss’s law, field between the sphere and the shell is determined by q1 alone. Hence, potential difference between the sphere and the shell is independent of q2. If q1 is positive, this potential difference is always positive. (a) Our body and the ground form an equipotential surface. As we step out into the open, the original equipotential surfaces of 2020-21

Physics open air change, keeping our head and the ground at the same potential. (b) Yes. The steady discharging current in the atmosphere charges up the aluminium sheet gradually and raises its voltage to an extent depending on the capacitance of the capacitor (formed by the sheet, slab and the ground). (c) The atmosphere is continually being charged by thunderstorms and lightning all over the globe and discharged through regions of ordinary weather. The two opposing currents are, on an average, in equilibrium. (d) Light energy involved in lightning; heat and sound energy in the accompanying thunder. CHAPTER 3 292 3.1 30 A 3.2 17 Ω, 8.5 V 3.3 (a) 6 Ω 3.4 (b) 2 V, 4 V, 6 V 3.5 (a) (20/19) Ω 3.6 3.7 (b) 10A, 5 A, 4A; 19A 3.8 3.9 1027 °C 2.0 × 10–7 Ω m 3.10 0.0039 °C–1 3.11 867 °C 3.12 3.13 Current in branch AB = (4/17) A, 3.14 in BC = (6/17) A, in CD = (–4/17) A, 3.15 in AD = (6/17) A, in BD. = (–2/17) A, total current = (10/17) A. 3.16 (a) X = 8.2 Ω; to minimise resistance of the connection which are 3.17 not accounted for in the bridge formula. (b) 60.5 cm from A. (c) The galvanometer will show no current. 11.5 V; the series resistor limits the current drawn from the external source. In its absence, the current will be dangerously high. 2.25 V 2.7 × 104 s (7.5 h) Take the radius of the earth = 6.37 × 106 m and obtain total charge of the globe. Divide it by current to obtain time = 283 s. Still this method gives you only an estimate; it is not strictly correct. Why? (a) 1.4 A, 11.9 V (b) 0.005 A; impossible because a starter motor requires large current ( ~ 100 A) for a few seconds. The mass (or weight) ratio of copper to aluminium wire is (1.72/2.63) × (8.9/2.7) ≅ 2.2. Since aluminium is lighter, it is preferred for long suspensions of cables. Ohm’s law is valid to a high accuracy; the resistivity of the alloy manganin is nearly independent of temperature. 2020-21

Answers 3.18 (a) Only current (because it is given to be steady!). The rest depends on the area of cross-section inversely. 3.19 3.20 (b) No, examples of non-ohmic elements: vacuum diode, semiconductor diode. (c) Because the maximum current drawn from a source = ε/r. (d) Because, if the circuit is shorted (accidentally), the current drawn will exceed safety limits, if internal resistance is not large. (a) greater, (b) lower, (c) nearly independent of, (d) 1022. (a) (i) in series, (ii) all in parallel; n2. (b) (i) Join 1 Ω, 2 Ω in parallel and the combination in series with 3Ω, (ii) parallel combination of 2 Ω and 3 Ω in series with 1 Ω, (iii) all in series, (iv) all in parallel. (c) (i) (16/3) Ω, (ii) 5 R. 3.21 Hint: Let X be the equivalent resistance of the infinite network. 3.22 Clearly, 2 + X/(X +1) = X which gives X = (1 + 3 ) Ω; therefore the current is 3.7 A. (a) ε = 1.25 V. (b) To reduce current through the galvanometer when the movable contact is far from the balance point. (c) No. (d) No. If ε is greater than the emf of the driver cell of the potentiometer, there will be no balance point on the wire AB. (e) The circuit, as it is, would be unsuitable, because the balance point (for ε of the order of a few mV) will be very close to the end A and the percentage error in measurement will be very large. The circuit is modified by putting a suitable resistor R in series with the wire AB so that potential drop across AB is only slightly greater than the emf to be measured. Then, the balance point will be at larger length of the wire and the percentage error will be much smaller. 3.23 1.7 Ω CHAPTER 4 293 4.1 π × 10–4 T 3.1 × 10–4 T 4.2 3.5 × 10–5 T 4.3 4 × 10–6 T, vertical up 4.4 1.2 × 10–5 T, towards south 4.5 0.6 N m–1 4.6 8.1 × 10–2 N; direction of force given by Fleming’s left-hand rule 4.7 2 × 10–5 N; attractive force normal to A towards B 2020-21

Physics 4.8 8π × 10–3 T 2.5 × 10–2 T 4.9 4.10 0.96 N m 4.11 4.12 (a) 1.4, (b) 1 4.13 4.2 cm 4.14 4.15 18 MHz 4.16 (a) 3.1 Nm, (b) No, the answer is unchanged because the formula τ = N I A × B is true for a planar loop of any shape. 5π × 10–4 T = 1.6 × 10–3 T towards west. Length about 50 cm, radius about 4 cm, number of turns about 400, current about 10 A. These particulars are not unique. Some adjustment with limits is possible. (b) In a small region of length 2d about the mid-point between the coils, µ0IR 2N  R d  2 R 2 −3/2  R − d  2  −3 / 2  2 2 2  B= × + + + + R 2  − µ0IR2N  5R2  −3 / 2 1 + 4d  −3 / 2 1 − 4d  −3 / 2  2  4  5R 5R  × × + µ0IR2N  4  3 / 2 1 − 6d 6d  2R3 5 5R 5R  − × × +1+ where in the second and third steps above, terms containing d2/R 2 and higher powers of d/R are neglected since d << 1 . The terms R linear in d/R cancel giving a uniform field B in a small region:  4  3 / 2 µ0IN − 0.72 µ0IN 5 R R B = 4.17 Hint: B for a toroid is given by the same formula as for a solenoid: 4.18 B = µ0 nI, where n in this case is given by n = N . The field is 4.19 2πr 4.20 non-zero only inside the core surrounded by the windings. (a) Zero, (b) 3.0 × 10–2 T, (c) zero. Note, the field varies slightly across the cross-section of the toroid as r varies from the inner to outer radius. Answer (b) corresponds to the mean radius r = 25.5 cm. (a) Initial v is either parallel or anti-parallel to B. (b) Yes, because magnetic force can change the direction of v, not its magnitude. (c) B should be in a vertically downward direction. (a) Circular trajectory of radius 1.0 mm normal to B. (b) Helical trajectory of radius 0.5 mm with velocity component 2.3 × 107 m s–1 along B. 294 Deuterium ions or deuterons; the answer is not unique because only the ratio of charge to mass is determined. Other possible answers are He++, Li+++ , etc. 2020-21

Answers 4.21 (a) A horizontal magnetic field of magnitude 0.26 T normal to the 4.22 conductor in such a direction that Fleming’s left-hand rule gives a magnetic force upward. 4.23 4.24 (b) 1.176 N. 4.25 1.2 N m–1; repulsive. Note, obtaining total force on the wire as 4.26 4.27 1.2 × 0.7 = 0.84 N, is only approximately correct because the formula 4.28 µ0 F = 2πr I1 I2 for force per unit length is strictly valid for infinitely long conductors. (a) 2.1 N vertically downwards (b) 2.1 N vertically downwards (true for any angle between current and direction and B since l sin θ remains fixed, equal to 20 cm) (c) 1.68 N vertically downwards Use τ = IA × B and F = I l × B (a) 1.8 × 10–2 N m along –y direction (b) same as in (a) (c) 1.8 × 10–2 N m along –x direction (d) 1.8 × 10–2 N m at an angle of 240° with the +x direction (e) zero (f ) zero Force is zero in each case. Case (e) corresponds to stable, and case (f ) corresponds to unstable equilibrium. (a) Zero, (b) zero, (c) force on each electron is evB = IB/(nA) = 5 × 10–25 N. Note: Answer (c) denotes only the magnetic force. 108 A Resistance in series = 5988 Ω Shunt resistance = 10 mΩ CHAPTER 5 5.1 (a) Magnetic declination, angle of dip, horizontal component of earth’s magnetic field. (b) Greater in Britain (it is about 70°), because Britain is closer to the magnetic north pole. (c) Field lines of B due to the earth’s magnetism would seem to come out of the ground. (d) A compass is free to move in a horizontal plane, while the earth’s field is exactly vertical at the magnetic poles. So the compass can point in any direction there. (e) Use the formula for field B on the normal bisector of a dipole of magnetic moment m, BA = − µ0 m 4π r3 Take m = 8 × 1022 J T–1, r = 6.4 × 106 m; one gets B = 0.3 G, which checks with the order of magnitude of the observed field on the earth. (f) Why not? The earth’s field is only approximately a dipole field. 295 Local N-S poles may arise due to, for instance, magnetised mineral deposits. 2020-21

Physics 5.2 (a) Yes, it does change with time. Time scale for appreciable change is roughly a few hundred years. But even on a much smaller 5.3 scale of a few years, its variations are not completely negligible. 5.4 5.5 (b) Because molten iron (which is the phase of the iron at the high 5.6 temperatures of the core) is not ferromagnetic. 5.7 5.8 (c) One possibility is the radioactivity in the interior of the earth. But nobody really knows. You should consult a good modern 5.9 text on geomagnetism for a proper view of the question. 5.10 5.11 (d) Earth’s magnetic field gets weakly ‘recorded’ in certain rocks 5.12 during solidification. Analysis of this rock magnetism offers 5.13 clues to geomagnetic history. 5.14 5.15 (e) At large distances, the field gets modified due to the field of ions in motion (in the earth’s ionosphere). The latter is sensitive to 296 extra-terrestrial disturbances such as, the solar wind. (f ) From the relation R = mv , an extremely minute field bends eB charged particles in a circle of very large radius. Over a small distance, the deflection due to the circular orbit of such large R may not be noticeable, but over the gigantic interstellar distances, the deflection can significantly affect the passage of charged particles, for example, cosmic rays. 0.36 J T–1 (a) m parallel to B; U = –mB = – 4.8 × 10–2 J: stable. (b) m anti-parallel to B; U = +mB = +4.8 × 10–2 J; unstable. 0.60 J T–1 along the axis of the solenoid determined by the sense of flow of the current. 7.5 ×10–2 J (a) (i) 0.33 J (ii) 0.66 J (b) (i) Torque of magnitude 0.33 J in a direction that tends to align the magnitude moment vector along B. (ii) Zero. (a) 1.28 A m2 along the axis in the direction related to the sense of current via the right-handed screw rule. (b) Force is zero in uniform field; torque = 0.048 Nm in a direction that tends to align the axis of the solenoid (i.e., its magnetic moment vector) along B. Use to get I = 1.2 ×10−4 kg m2. B = 0.35 sec 22° 0.38 G. The earth’s lies in the vertical plane 12° west of the geographic meridian making an angle of 60° (upwards) with the horizontal (magnetic south to magnetic north) direction. Magnitude = 0.32 G. (a) 0.96 g along S-N direction. (b) 0.48 G along N-S direction. 0.54 G in the direction of earth’s field. At 14 × 2–1/3 = 11.1 cm on the normal bisector. (a) (µ0m )/(4πr 3 ) = 0.42 × 10−4 which gives r = 5.0 cm. (b) (2µ0m )/(4πr13 ) = 0.42 × 10−4 i.e., r1 = 21/3 r = 6.3 cm. 2020-21

Answers 5.16 (a) The tendency to disrupt the alignment of dipoles (with the magnetising field) arising from random thermal motion is 5.17 reduced at lower temperatures. 5.18 (b) The induced dipole moment in a diamagnetic sample is always 5.19 opposite to the magnetising field, no matter what the internal motion of the atoms is. (c) Slightly less, since bismuth is diamagnetic. (d) No, as it evident from the magnetisation curve. From the slope of magnetisation curve, it is clear that m is greater for lower fields. (e) Proof of this important fact (of much practical use) is based on boundary conditions of magnetic fields (B and H) at the interface of two media. (When one of the media has µ >> 1, the field lines meet this medium nearly normally.) Details are beyond the scope of this book. (f ) Yes. Apart from minor differences in strength of the individual atomic dipoles of two different materials, a paramagnetic sample with saturated magnetisation will have the same order of magnetisation. But of course, saturation requires impractically high magnetising fields. (b) Carbon steel piece, because heat lost per cycle is proportional to the area of hysteresis loop. (c) Magnetisation of a ferromagnet is not a single-valued function of the magnetising field. Its value for a particular field depends both on the field and also on history of magnetisation (i.e., how many cycles of magnetisation it has gone through, etc.). In other words, the value of magnetisation is a record or memory of its cycles of magnetisation. If information bits can be made to correspond to these cycles, the system displaying such a hysteresis loop can act as a device for storing information. (d) Ceramics (specially treated barium iron oxides) also called ferrites. (e) Surround the region by soft iron rings. Magnetic field lines will be drawn into the rings, and the enclosed space will be free of magnetic field. But this shielding is only approximate, unlike the perfect electric shielding of a cavity in a conductor placed in an external electric field. Parallel to and above the cable at a distance at 1.5 cm. Below the cable: Rh = 0.39 cos35° – 0.2 = 0.12 G Rv = 0.36 sin35° = 0.22 G R= R 2 + Rv2 = 0.25 G h θ = tan−1 Rv = 62° Rh Above the cable: Rh = 0.39 cos35° + 0.2 = 0.52 G Rv = 0.224 G 297 R = 0.57 G, θ 23° 2020-21

Physics 5.20 (a) Bh = (µ0IN /2r ) cos 45o = 0.39 G 5.21 (b) East to west (i.e., the needle will reverse its original direction). Magnitude of the other field = 1.2 × 10−2 × sin15o sin 45o = 4.4 × 10−3 T 5.22 R = meV eB = 2me × kinetic energy eB = 11.3 m Up or down deflection = R (1-cosθ ) where sinθ = 0.3/11.3. We get deflection 4 mm. 5.23 Initially, total dipole moment 5.24 = 0.15 × 1.5 × 10-23 × 2.0 ×1024 5.25 = 4.5 J T –1 Use Curie’s Law m ∝ B/T to get the final dipole moment = 4.5 × (0.98/0.84) × (4.2/2.8) = 7.9 J T –1 Use the formula B = µr µo NI where µr (relative permeability) to get 2π R B = 4.48 T. Of the two, ftohlleowreslaetaiosnilyµlfr=om– (eth/2emd)elfiinsitiinonasccoofrdµal nanced with classical physics. It l: µl = IA = (e / T )πr 2 l = mvr = m 2πr 2 T where r is the radius of the circular orbit which the electron of mass m and charge (–e) completes in time T. Clearly, µr /l = e / 2m. Since charge of the electron is negative (= – e), it is easily seen that µ and l are antiparallel, both normal to the plane of the orbit. Therefore, µl = −e / 2m l. Note µs/S in contrast to µl /l is e/m, i.e., twice the classically expected value. This latter result (verified experimentally) is an outstanding consequence of modern quantum theory and cannot be obtained classically. 298 CHAPTER 6 6.1 (a) Along qrpq (b) Along prq, along yzx 2020-21

Answers (c) Along yzx (d) Along zyx (e) Along xry (f ) No induced current since field lines lie in the plane of the loop. 6.2 (a) Along adcd (flux through the surface increases during shape 6.3 change, so induced current produces opposing flux). 6.4 6.5 (b) Along a′d′c′b′ (flux decreases during the process) 6.6 7.5 × 10–6 V (1) 2.4 × 10–4 V, lasting 2 s 6.7 (2) 0.6 × 10–4 V, lasting 8 s 6.8 6.9 100 V 6.10 Flux through each turn of the loop = π r 2B cos(ωt) 6.11 ε = –N ω π r 2B sin(ωt) εmax = –N ω π r 2B 6.12 = 20 × 50 × π × 64 × 10–4 × 3.0 × 10–2 = 0.603 V εavg is zero over a cycle Imax = 0.0603 A Paverage = 1 ε max I max = 0.018 W 2 The induced current causes a torque opposing the rotation of the coil. An external agent (rotor) must supply torque (and do work) to counter this torque in order to keep the coil rotating uniformly. Thus, the source of the power dissipated as heat in the coil is the external rotor. (a) 1.5 × 10–3 V, (b) West to East, (c) Eastern end. 4H 30 Wb Vertical component of B = 5.0 × 10 –4 sin 30° = 2.5 × 10 –4 T ε = Blv ε = 2.5 × 10 –4 × 25 × 500 = 3.125 V The emf induced is 3.1 V (using significant figures). The direction of the wing is immaterial (as long as it is horizontal) for this answer. Induced emf = 8 × 2 × 10 –4 × 0.02 = 3.2 × 10–5 V Induced current = 2 × 10 –5 A Power loss = 6.4 × 10 –10 W Source of this power is the external agent responsible for changing the magnetic field with time. Rate of change of flux due to explicit time variation in B = 144 × 10 –4 m2 × 10–3 T s–1 = 1.44 × 10 –5 Wb s–1 Rate of change of flux due to motion of the loop in a non-uniform B 299 = 144 × 10 –4 m2 × 10–3 T cm–1 × 8 cm s–1 = 11.52 × 10–5 Wb s–1 2020-21

Physics The two effects add up since both cause a decrease in flux along the positive z-direction. Therefore, induced emf = 12.96 × 10–5 V; induced current = 2.88 × 10–2 A. The direction of induced current is such as to increase the flux through the loop along positive z-direction. If for the observer the loop moves to the right, the current will be seen to be anti-clockwise. A proper proof of the procedure above is as follows: a Φ(t ) = ∫ aB (x,t )dx 0 ∫dΦ = a dB(x,t ) a dx dt 0 dt using, dB = ∂B + ∂B dx dt ∂t ∂x dt  ∂B + v ∂B  =  ∂t ∂x  we get, ∫dΦ a  ∂B(x ,t ) + v ∂B(x,t )  ∂t ∂x  dt = a dx 0 = A  ∂B + v ∂B   ∂t ∂x  where A = a2 The last step follows because  ∂B  ,  ∂B  and v are given to be ∂t ∂x constants in the problem. Even if you do not understand this formal proof (which requires good familiarity with calculus), you will still appreciate that flux change can occur both due to the motion of the loop as well as time variations in the magnetic field. 6.13 tf Q = ∫ Idt ti ∫= 1 tf ε dt R ti ∫= − N Φf dΦ R Φi = N (Φi − Φf ) R for N = 25, R = 0.50 Ω, Q = 7.5 × 10–3 C 300 Φf = 0, A = 2.0 × 10–4 m2, Φi = 1.5 × 10–4 Wb B = Φi/A = 0.75 T 2020-21

Answers 6.14 |ε | = vBl = 0.12 × 0.50 × 0.15 = 9.0 mV; P positive end and Q negative end. (b) Yes. When K is closed, the excess charge is maintained by the continuous flow of current. (c) Magnetic force is cancelled by the electric force set-up due to the excess charge of opposite signs at the ends of the rod. (d) Retarding force = I B l = 9 mV × 0.5 T × 0.15 m 9 mΩ = 75 × 10–3 N (e) Power expended by an external agent against the above retarding force to keep the rod moving uniformly at 12 cm s–1 = 75 × 10–3 × 12 × 10–2 = 9.0 × 10–3 W When K is open, no power is expended. ( f ) I2R = 1 × 1 × 9 × 10–3 = 9.0 × 10–3 W The source of this power is the power provided by the external agent as calculated above. (g) Zero; motion of the rod does not cut across the field lines. [Note: length of PQ has been considered above to be equal to the spacing between the rails.] 6.15 B = µ0NI l (Inside the solenoid away from the ends) Φ = µ0NI A l Total flux linkage = NΦ = µ0N 2 A I l (Ignoring end variations in B) ε = d (NΦ) dt ε = total change in flux av total time ε = 4π × 10−7 × 25 × 10−4 × (500)2 × 2.5 av 0.3 × 10−3 = 6.5 V 6.16 M = µ0 a ln 1 + a  2π x ε = 1.7 × 10–5 V 6.17 − B π a 2 λ kˆ 301 MR 2020-21

Physics CHAPTER 7 7.1 (a) 2.20 A (b) 484 W 7.2 (a) 300 = 212.1 V 2 (b) 10 2 =14.1 A 7.3 15.9 A 7.4 2.49 A 7.5 Zero in each case. 7.6 125 s–1; 25 7.7 1.1 × 103 s–1 7.8 0.6 J, same at later times. 7.9 2,000 W 7.10 ν = 1 1 , i.e., C = 1 7.11 2π LC 4π2ν2L 7.12 For L = 200 µH, ν = 1200 kHz, C = 87.9 pF. For L = 200 µH, ν = 800 kHz, C = 197.8 pF. The variable capacitor should have a range of about 88 pF to 198 pF. (a) 50 rad s–1 (b) 40 Ω, 8.1 A (c) VLrms = 1437.5 V, VCrms = 1437.5 V , VRrms = 230 V VLCrms =  − 1 = 0 Irms  ω0L ω0 C  (a) 1.0 J. Yes, sum of the energies stored in L and C is conserved if R = 0. (b) ω = 103 rad s−1, ν = 159 Hz (c) q = q0 cos ω t (i) Energy stored is completely electrical at t = 0, T , T , 3T , ..... 22 (ii) Energy stored is completely magnetic (i.e., electrical energy is zero) at t = T, 3T , 5T ....... , where T = 1 = 6.3 ms . 4 4 4 ν (d) At t = T, 3T , 5T ,....... , because q = q 0 cos ωT = q0 cos π = q0 . 8 8 8 8 4 2 Therefore, electrical energy = q2 = 1  q2  which is half the total 2C  0  2 2C energy. 302 (e) R damps out the LC oscillations eventually. The whole of the initial energy (= 1.0 J) is eventually dissipated as heat. 2020-21

Answers 7.13 For an LR circuit, if V = V0 sin ω t V I = 0 sin (ω t − φ ), where tan φ = (ω L / R) . R2 + ω 2L2 (a) I0 = 1.82 A (b) V is maximum at t = 0, I is maximum at t = (φ / ω ). Now, tan φ = 2 π ν L = 1.571 or φ ≈ 57.5° R Therefore, time lag = 57.5 π × 2π 1 50 = 3.2 ms 180 × 7.14 (a) I 0 = 1.1×10−2 A 7.15 (b) tan φ = 100 π, φ is close to π/2. 7.16 I0 is much smaller than the low frequency case (Exercise 7.13) 7.17 showing thereby that at high frequencies, L nearly amounts to an open circuit. In a dc circuit (after steady state) ω = 0, so here L acts like a pure conductor. For a RC circuit, if V = V0 sin ω t I = V0 sin(ω t + φ ) where tan φ = 1 R ωC R2 + (1 / ω C )2 (a) I0 = 3.23 A (b) φ = 33.5° Time lag = φ = 1.55 ms ω (a) I0 = 3.88 A (b) φ ≈ 0.2 and is nearly zero at high frequency. Thus, at high frequency, C acts like a conductor. For a dc circuit, after steady state, ω = 0 and C amounts to an open circuit. Effective impedance of the parallel LCR circuit is given by 1= 1 +  − 1 2 Z R2 ω C ω L  which is minimum at ω = ω0 = 1 LC Therefore, |Z| is maximum at ω = ω0 , and the total current amplitude is minimum. In R branch, IRrms = 5.75 A In L branch, ILrms = 0.92 A In C branch, ICrms = 0.92 A 7.18 Note: total current Irms = 5.75 A , since the currents in L and C branch 303 are 180° out of phase and add to zero at every instant of the cycle. (a) For V = V0 sin ω t 2020-21

Physics I = V0 sin  + π  if R = 0 ωL − 1 ωt 2 ; ωC where – sign appears if ωL > 1/ωC, and + sign appears if ωL < 1/ωC. I0 = 11.6 A, Irms = 8.24 A (b) VL rms = 207 V,VCrms = 437 V (Note: 437 V – 207 V = 230 V is equal to the applied rms voltage as should be the case. The voltage across L and C gets subtracted because they are 180° out of phase.) (c) Whatever be the current I in L, actual voltage leads current by π/2. Therefore, average power consumed by L is zero. (d) For C, voltage lags by π/2. Again, average power consumed by C is zero. (e) Total average power absorbed is zero. 7.19 Irms = 7.26 A Average power to R = I 2 R = 791 W rms Average power to L = Average power to C = 0 Total power absorbed = 791 W 7.20 (a) ω0= 4167 rad s–1; ν0 = 663 Hz I max = 14.1 A 0 (b) P = (1 / 2) I 2 R which is maximum at the same frequency (663 Hz) 0 for which I0 is maximum Pmax = (1/ 2)(I max )2 R = 2300 W. (c) At ω = ω0 ± ∆ω [Approximation good if (R/2L ) << ω0 ]. ∆ω = R/2L = 95.8 rad s–1; ∆ν = ∆ω/2π = 15.2 Hz. Power absorbed is half the peak power at ν = 648 Hz and 678 Hz. At these frequencies, current amplitude is (1/ 2) times I max , 0 i.e., current amplitude(at half the peak power points) is 10 A. 7.21 (d) Q = 21.7 7.22 ω0= 111 rad s–1; Q = 45 To double Q without changing ω0, reduce R to 3.7 Ω. (a) Yes. The same is not true for rms voltage, because voltages across different elements may not be in phase. See, for example, answer to Exercise 7.18. (b) The high induced voltage, when the circuit is broken, is used to charge the capacitor, thus avoiding sparks, etc. (c) For dc, impedance of L is negligible and of C very high (infinite), so the dc signal appears across C. For high frequency ac, impedance of L is high and that of C is low. So, the ac signal appears across L. (d) For a steady state dc, L has no effect, even if it is increased by an iron core. For ac, the lamp will shine dimly because of additional impedance of the choke. It will dim further when the iron core is inserted which increases the choke’s impedance. 304 (e) A choke coil reduces voltage across the tube without wasting power. A resistor would waste power as heat. 2020-21

Answers 7.23 400 7.24 Hydroelectric power = h ρ g × A × v = h ρ g β 7.25 where β = Av is the flow (volume of water flowing per second across a cross-section). 7.26 Electric power available = 0.6 × 300 × 103 × 9.8 × 100 W = 176 MW Line resistance = 30 × 0.5 = 15 Ω. rms current in the line = 800 ×1000 W = 200 A 4000 V (a) Line power loss = (200 A)2 × 15 Ω = 600 kW. (b) Power supply by the plant = 800 kW + 600 kW = 1400 kW. (c) Voltage drop on the line = 200 A × 15 Ω = 3000 V. The step-up transformer at the plant is 440 V – 7000 V. Current = 800 ×1000 W = 20 A 40,000 V (a) Line power loss = (20 A)2 × (15 Ω) = 6 kW. (b) Power supply by the plant = 800 kW + 6 kW = 806 kW. (c) Voltage drop on the line = 20 A × 15 Ω = 300 V. The step-up transformer is 440 V – 40, 300 V. It is clear that percentage power loss is greatly reduced by high voltage transmission. In Exercise 7.25, this power loss is (600/1400) × 100 = 43%. In this exercise, it is only (6/806) × 100 = 0.74%. CHAPTER 8 8.1 (a) C = ε0 A /d = 80.1 pF dQ = C dV dt dt dV = 0.15 = 1.87 × 109 V s–1 dt 80.1 × 10–12 (b) id = ε0 d ΦΕ. . Now across the capacitor ΦE = EA, ignoring end dt corrections. Therefore, id = ε0 A dΦΕ dt Now, E = Q . Therefore, dE = i , which implies id = i = 0.15 A. εA dt εA 00 (c) Yes, provided by ‘current’ we mean the sum of conduction and 305 displacement currents. 8.2 (a) Irms = Vrms ωC = 6.9µA 2020-21

Physics (b) Yes. The derivation in Exercise 8.1(b) is true even if i is oscillating in time. (c) The formula B = µ0 r id 2π R2 goes through even if id (and therefore B ) oscillates in time. The formula shows they oscillate in phase. Since id = i, we have B0 and i0 are the amplitudes of the oscillating B0 = µ0 r i0 , where 2π R2 magnetic field and current, respectively. i0= 2I = 9.76 µA. For r = 3 cm, R = 6 cm, B0 = 1.63 × 10–11 T. rms 8.3 The speed in vacuum is the same for all: c = 3 × 108 m s–1. 8.4 E and B in x-y plane and are mutually perpendicular, 10 m. 8.5 Wavelength band: 40 m – 25 m. 8.6 109 Hz 8.7 153 N/C 8.8 (a) 400 nT, 3.14 × 108 rad/s, 1.05 rad/m, 6.00 m. (b) E = { (120 N/C) sin[(1.05 rad/m)]x – (3.14 × 108 rad/s)t]} ˆj B = { (400 nT) sin[(1.05 rad/m)]x – (3.14 × 108 rad/s)t ]} kˆ 8.9 Photon energy (for λ = 1 m) 6.63 × 10−34 × 3 × 108 eV = 1.24 × 10−6 eV = 1.6 × 10−19 Photon energy for other wavelengths in the figure for electromagnetic spectrum can be obtained by multiplying approximate powers of ten. Energy of a photon that a source produces indicates the spacings of the relevant energy levels of the source. For example, λ = 10–12 m corresponds to photon energy = 1.24 × 106 eV = 1.24 MeV. This indicates that nuclear energy levels (transition between which causes γ-ray emission) are typically spaced by 1 MeV or so. Similarly, a visible wavelength λ = 5 × 10–7 m, corresponds to photon energy = 2.5 eV. This implies that energy levels (transition between which gives visible radiation) are typically spaced by a few eV. 8.10 (a) λ = (c/ν) = 1.5 × 10–2 m (b) B0 = (E0/c) = 1.6 × 10–7 T (c) Energy density in E field: uE = (1/2)ε0 E 2 Energy density in B field: uB = (1/2µ0)B 2 1 Using E = cB, and c = µ0ε0 , uE = uB 8.11 (a) – ˆj , (b) 3.5 m, (c) 86 MHz, (d) 100 nT, (e) {(100 nT) cos[(1.8 rad/m)y + (5.4 × 106 rad/s)t]} kˆ 306 8.12 (a) 0.4 W/m2, (b) 0.004 W/m2 2020-21

Answers 8.13 A body at temperature T produces a continuous spectrum of wavelengths. For a black body, the wavelength corresponding to 8.14 maximum intensity of radiation is given according to Wein’s law by 8.15 the relation: λm=0.29 cm K/T. For λm=10–6 m, T = 2900 K. Temperatures for other wavelengths can be found. These numbers tell us the temperature ranges required for obtaining radiations in different parts of the electromagnetic spectrum. Thus, to obtain visible radiation, say λ = 5 × 10–7 m, the source should have a temperature of about 6000 K. Note: a lower temperature will also produce this wavelength but not the maximum intensity. (a) Radio (short wavelength end) (b) Radio (short wavelength end) (c) Microwave (d) Visible ( Yellow ) (e) X-rays (or soft γ-rays) region (a) Ionosphere reflects waves in these bands. (b) Television signals are not properly reflected by the ionosphere (see text). Therefore, reflection is effected by satellites. (c) Atmosphere absorbs X-rays, while visible and radiowaves can penetrate it. (d) It absorbs ultraviolet radiations from the sun and prevents it from reaching the earth’s surface and causing damage to life. (e) The temperature of the earth would be lower because the Greenhouse effect of the atmosphere would be absent. ( f ) The clouds produced by global nuclear war would perhaps cover substantial parts of the sky preventing solar light from reaching many parts of the globe. This would cause a ‘winter’. 307 2020-21

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Physics ANSWERS CHAPTER 9 9.1 v = –54 cm. The image is real, inverted and magnified. The size of the 9.2 image is 5.0 cm. As u → f, v → ∞; for u < f, image is virtual. 9.3 9.4 v = 6.7 cm. Magnification = 5/9, i.e., the size of the image is 2.5cm. As 9.5 u → ∞; v → f (but never beyond) while m → 0. 9.6 9.7 1.33; 1.7 cm 9.8 nga = 1.51; n = 1.32; ngw = 1.144; which gives sin r = 0.6181 i.e., 9.9 wa 9.10 9.11 r ~ 38°. 9.12 r = 0.8 × tan ic and sin ic = 1/1.33 ≅ 0.75 , where r is the radius (in m) of the largest circle from which light comes out and ic is the critical angle for water-air interface, Area = 2.6 m2 n ≅ 1.53 and Dm for prism in water ≅ 10° R = 22cm Here the object is virtual and the image is real. u = +12 cm (object on right; virtual) (a) f = +20 cm. Image is real and at 7.5 cm from the lens on its right side. (b) f = –16 cm. Image is real and at 48 cm from the lens on its right side. v = 8.4 cm, image is erect and virtual. It is diminished to a size 1.8 cm. As u → ∞, v → f (but never beyond f while m → 0). Note that when the object is placed at the focus of the concave lens (21 cm), the image is located at 10.5 cm (not at infinity as one might wrongly think). A diverging lens of focal length 60 cm (a) v1e0=cm–2, 5 cm and fe = 6.25 cm give ue = –5 cm; vO = (15 – 5) cm = fO = uO = – 2.5 cm; Magnifying power = 20 (b) uO = – 2.59 cm. Magnifying power = 13.5. Angular magnification of the eye-piece for image at 25 cm = 25 + 1 = 11; | ue |= 25 cm = 2.27cm ; vO = 7.2 cm 2.5 11 502 Separation = 9.47 cm; Magnifying power = 88 2020-21

Answers 9.13 24; 150 cm 9.14 9.15 (a) Angular magnification = 1500 9.16 9.17 (b) Diameter of the image = 13.7 cm. 9.18 Apply mirror equation and the condition: 9.19 (a) f < 0 (concave mirror); u < 0 (object on left) 9.20 (b) f > 0; u < 0 (c) f > 0 (convex mirror) and u < 0 (d) f < 0 (concave mirror); f < u < 0 to deduce the desired result. The pin appears raised by 5.0 cm. It can be seen with an explicit ray diagram that the answer is independent of the location of the slab (for small angles of incidence). (a) sin eic′s= 1.44/1.68 which gives oicr′ = 59°. Total internal reflection tak place when i > 59° when r < rmax = 31°. Now, (sinimax /sinrmax ) = 1.68 , which gives imax ~ 60°. Thus, all incident rays of angles in the range 0 < i < 60° will suffer total internal reflections in the pipe. (If the length of the pipe is finite, which it is in practice, there will be a lower limit on i determined by the ratio of the diameter to the length of the pipe.) (b) If there is no outer coating, ic′ = sin–1(1/1.68) = 36.5°. Now, i = 90° will have r = 36.5° and i ′ = 53.5° which is greater than isc′u. fTfehrutso,taalllinintecridneanl trerfalyescti(oinnst.he range 53.5° < i < 90°) will (a) Rays converging to a point ‘behind’ a plane or convex mirror are reflected to a point in front of the mirror on a screen. In other words, a plane or convex mirror can produce a real image if the object is virtual. Convince yourself by drawing an appropriate ray diagram. (b) When the reflected or refracted rays are divergent, the image is virtual. The divergent rays can be converged on to a screen by means of an appropriate converging lens. The convex lens of the eye does just that. The virtual image here serves as an object for the lens to produce a real image. Note, the screen here is not located at the position of the virtual image. There is no contra- diction. (c) Taller (d) The apparent depth for oblique viewing decreases from its value for near-normal viewing. Convince yourself of this fact by draw- ing ray diagrams for different positions of the observer. (e) Refractive index of a diamond is about 2.42, much larger than that of ordinary glass (about 1.5). The critical angle of diamond is about 24°, much less than that of glass. A skilled diamond- cutter exploits the larger range of angles of incidence (in the diamond), 24° to 90°, to ensure that light entering the diamond is totally reflected from many faces before getting out–thus producing a sparkling effect. For fixed distance s between object and screen, the lens equation does not give a real solution for u or v if f is greater than s/4. Therefore, fmax = 0.75 m. 503 21.4 cm 2020-21

Physics 9.21 (a) (i) Let a parallel beam be the incident from the left on the convex lens first. 9.22 9.23 f1 = 30 cm and u1 = – ∞ , give v1 = + 30 cm. This image becomes a virtual object for the second lens. f2 = –20 cm, u2 = + (30 – 8) cm = + 22 cm which gives, fvr2om= – 220 cm. The parallel incident beam appears to diverge a point 216 cm from the centre of the two-lens system. (ii) Let the parallel beam be incident from the left on the concave lens first: f1 = – 20 cm, u1 = – ∞, give v1 = – 20 cm. This image becomes a real object for the second =le–ns4:2f02 =+ T3h0ecpma,rua2lle=l – (20 + 8) cm = – 28 cm which gives, v2 cm. incident beam appears to diverge from a point 416 cm on the left of the centre of the two-lens system. Clearly, the answer depends on which side of the lens system the parallel beam is incident. Further we do not have a simple lens equation true for all u (and v) in terms of a definite constant of the system (the constant bTehinegndoettieornmionfeedffbeyctfi1vae nfdocfa2,l and the separation between the lenses). length, therefore, does not seem to be meaningful for this system. (b) u1 = – 40 cm, f1 = 30 cm, gives v1= 120 cm. Magnitude of magnification due to the first (convex) lens is 3. u 2 = + (120 – 8) cm = +112 cm (object virtual); f2 = – 20 cm which gives v2 = − 112 × 20 cm 92 Magnitude of magnification due to the second (concave) lens = 20/92. Net magnitude of magnification = 0.652 Size of the image = 0.98 cm If the refracted ray in the prism is incident on the second face at the critical angle ic, the angle of refraction r at the first face is (60°–ic). Now, ic = sin–1 (1/1.524) ~ 41° Therefore, r = 19° sin i = 0.4962; i ~ 30° (a) 1 + 1 = 1 v 9 10 i.e., v = – 90 cm, Magnitude of magnification = 90/9 = 10. Each square in the virtual image has an area 10 × 10 × 1 mm2 = 100 mm2 = 1 cm2 (b) Magnifying power = 25/9 = 2.8 (c) No, magnification of an image by a lens and angular magnification (or magnifying power) of an optical instrument are two separate things. The latter is the ratio of the angular size of the object (which is equal to the angular size of the image even if the image is magnified) to the angular size of the object if placed at the near point (25 cm). Thus, magnification magnitude is |(v/u )| and magnifying power is (25/ |u|). Only when the image is located at the near point |v| = 25 cm, are the two quantities equal. 504 2020-21

Answers 9.24 (a) Maximum magnifying power is obtained when the image is at 9.25 the near point (25 cm) 9.26 u = – 7.14 cm. (b) Magnitude of magnification = (25/ |u|) = 3.5. (c) Magnifying power = 3.5 Yes, the magnifying power (when the image is produced at 25 cm) is equal to the magnitude of magnification. Magnification = (6.25 /1) = 2.5 v = +2.5u 1 11 + −= 2.5u u 10 i.e.,u = – 6 cm |v| = 15 cm The virtual image is closer than the normal near point (25 cm) and cannot be seen by the eye distinctly. (a) Even though the absolute image size is bigger than the object size, the angular size of the image is equal to the angular size of the object. The magnifier helps in the following way: without it object would be placed no closer than 25 cm; with it the object can be placed much closer. The closer object has larger angular size than the same object at 25 cm. It is in this sense that angular magnification is achieved. (b) Yes, it decreases a little because the angle subtended at the eye is then slightly less than the angle subtended at the lens. The effect is negligible if the image is at a very large distance away. [Note: When the eye is separated from the lens, the angles subtended at the eye by the first object and its image are not equal.] (c) First, grinding lens of very small focal length is not easy. More important, if you decrease focal length, aberrations ( both spherical and chromatic ) become more pronounced. So, in practice, you cannot get a magnifying power of more than 3 or so with a simple convex lens. However, using an aberration corrected lens system, one can increase this limit by a factor of 10 or so. (d) Angular magnification of eye-piece ims a[(g2n5i/fifcea) t+io1n] (offe in cm) which increases if fe is smaller. Further, the objective vO = 1 is given by |uO | (|uO |/ fO ) − 1 which is large when |u O| is slightly greater than fO. The micro- scope is used for viewing very close object. So |uO| is small, and so is fO. (e) The image of the objective in the eye-piece is known as ‘eye-ring’. All the rays from the object refracted by objective go through the eye-ring. Therefore, it is an ideal position for our eyes for viewing. If we place our eyes too close to the eye-piece, we shall not collect much of the light and also reduce our field of view. If we position 505 2020-21

Physics 9.27 our eyes on the eye-ring and the area of the pupil of our eye is 9.28 greater or equal to the area of the eye-ring, our eyes will collect all the light refracted by the objective. The precise location of the eye-ring naturally depends on the separation between the objective and the eye-piece. When you view through a microscope by placing your eyes on one end,the ideal distance between the eyes and eye-piece is usually built-in the design of the instrument. Assume microscope in normal use i.e., image at 25 cm. Angular magnification of the eye-piece = 25 +1 = 6 5 Magnification of the objective = 30 = 5 6 11 1 − = 1.25 5u O uO which gives uO= –1.5 cm; v0= 7.5 cm. |ue|= (25/6) cm = 4.17 cm. The separation between the objective and the eye-piece should be (7.5 + 4.17) cm = 11.67 cm. Further the object should be placed 1.5 cm from the objective to obtain the desired magnification. (a) m = ( fO/fe ) = 28 (b) m= fO 1 + fO  = 33.6 fe 25  9.29 (a) fO + fe = 145 cm (b) Angle subtended by the tower = (100/3000) = (1/30) rad. Angle subtended by the image produced by the objective = h = h fO 140 Equating the two, h = 4.7 cm. (c) Magnification (magnitude) of the eye-piece = 6. Height of the final image (magnitude) = 28 cm. 9.30 The image formed by the larger (concave) mirror acts as virtual object 9.31 for the smaller (convex) mirror. Parallel rays coming from the object 9.32 at infinity will focus at a distance of 110 mm from the larger mirror. The distance of virtual object for the smaller mirror = (110 – 20) = 90 mm. The focal length of smaller mirror is 70 mm. Using the mirror formula, image is formed at 315 mm from the smaller mirror. The reflected rays get deflected by twice the angle of rotation of the mirror. Therefore, d/1.5 = tan 7°. Hence d = 18.4 cm. n = 1.33 506 2020-21

Answers CHAPTER 10 10.1 (a) Reflected light: (wavelength, frequency, speed same as incident light) 10.2 10.3 λ = 589 nm, ν = 5.09 × 1014 Hz, c = 3.00 × 108 m s–1 (b) Refracted light: (frequency same as the incident frequency) ν = 5.09 × 1014Hz v = (c/n) = 2.26 × 108 m s–1, λ = (v/ν ) = 444 nm (a) Spherical (b) Plane (c) Plane (a small area on the surface of a large sphere is nearly planar). (a) 2.0 × 108 m s–1 (b) No. The refractive index, and hence the speed of light in a medium, depends on wavelength. [When no particular wavelength or colour of light is specified, we may take the given refractive index to refer to yellow colour.] Now we know violet colour deviates more than red in a glass prism, i.e. nv > nr. Therefore, the violet component of white light travels slower than the red component. 10.4 λ = 1.2 ×10–2 × 0.28×10– 3 m = 600 nm 4 ×1.4 10.5 K/4 10.6 (a) 1.17 mm (b) 1.56 mm 10.7 0.15° 10.8 tan–1(1.5) ~ 56.3o 10.9 5000 Å, 6 × 1014 Hz; 45° 10.10 40 m 10.11 Use the formula λ′ – λ = v λ c i.e., v = c (λ' − λ) = 3 ×108 ×15 = 6.86 × 105 m s–1 λ 6563 10.12 In corpuscular (particle) picture of refraction, particles of light incident from a rarer to a denser medium experience a force of attraction normal to the surface. This results in an increase in the normal component of the velocity but the component along the surface is unchanged. This means c sin i = v sin r or v sin i = n. Since n > 1, v > c. = c sin r The prediction is opposite to the experimental results (v < c ). The wave picture of light is consistent with the experiment. 507 2020-21

508 Physics 10.13 With the point object at the centre, draw a circle touching the mirror. This is a plane section of the spherical wavefront from the object that has just reached the mirror. Next draw the locations of this same wavefront after a time t in the presence of the mirror, and in the absence of the mirror. You will get two arcs symmetrically located on either side of the mirror. Using simple geometry, the centre of the reflected wavefront (the image of the object) is seen to be at the same distance from the mirror as the object. 10.14 (a) The speed of light in vacuum is a universal constant independent of all the factors listed and anything else. In particular, note the surprising fact that it is independent of the relative motion between the source and the observer. This fact is a basic axiom of Einstein’s special theory of relativity. (b) Dependence of the speed of light in a medium: (i) does not depend on the nature of the source (wave speed is determined by the properties of the medium of propagation. This is also true for other waves, e.g., sound waves, water waves, etc.). (ii) independent of the direction of propagation for isotropic media. (iii) independent of the motion of the source relative to the medium but depends on the motion of the observer relative to the medium. (iv) depends on wavelength. (v) independent of intensity. [For high intensity beams, however, the situation is more complicated and need not concern us here.] 10.15 Sound waves require a medium for propagation. Thus even though the situations (i) and (ii) may correspond to the same relative motion (between the source and the observer), they are not identical physically since the motion of the observer relative to the medium is different in the two situations. Therefore, we cannot expect Doppler formulas for sound to be identical for (i) and (ii). For light waves in vacuum, there is clearly nothing to distinguish between (i) and (ii). Here only the relative motion between the source and the observer counts and the relativistic Doppler formula is the same for (i) and (ii). For light propagation in a medium, once again like for sound waves, the two situations are not identical and we should expect the Doppler formulas for this case to be different for the two situations (i) and (ii). 10.16 3.4 × 10 – 4 m. 10.17 (a) The size reduces by half according to the relation: size ~ λ/d. Intensity increases four fold. (b) The intensity of interference fringes in a double-slit arrangement is modulated by the diffraction pattern of each slit. (c) Waves diffracted from the edge of the circular obstacle interfere constructively at the centre of the shadow producing a bright spot. (d) For diffraction or bending of waves by obstacles/apertures by a large angle, the size of the latter should be comparable to wavelength. If the size of the obstacle/aperture is much too large compared to wavelength, diffraction is by a small angle. Here the size is of the order of a few metres. The wavelength of light is about 5 × 10–7 m, while sound waves of, say, 1 kHz frequency 2020-21

Answers have wavelength of about 0.3 m. Thus, sound waves can bend around the partition while light waves cannot. (e) Justification based on what is explained in (d). Typical sizes of apertures involved in ordinary optical instruments are much larger than the wavelength of light. 10.18 12.5 cm. 10.19 0.2 nm. 10.20 (a) Interference of the direct signal received by the antenna with (b) the (weak) signal reflected by the passing aircraft. Superposition principle follows from the linear character of the (differential) equation governing wave motion. Icfoym1 bainndatyio2naroef solutions of the wave equation, so is any linear yb1eaamnds)ya2.nWd hneonnt-hlieneaamrpelfifteucdtsesaarreeimlarpgoert(ea.ngt.,, high intensity laser the situation is far more complicated and need not concern us here. 10.21 Divide the single slit into n smaller slits of width a ′ = a/n . The angle θ = n λ/a = λ/a ′. Each of the smaller slits sends zero intensity in the direction θ. The combination gives zero intensity as well. CHAPTER 11 11.1 (a) 7.24 × 1018 Hz (b) 0.041 nm 11.2 (a) 0.34 eV = 0.54 × 10–19J (b) 0.34 V (c) 344 km/s 11.3 1.5 eV = 2.4 × 10–19 J 11.4 (a) 3.14 × 10–19J, 1.05 × 10–27 kg m/s (b) 3 × 1016 photons/s (c) 0.63 m/s 11.5 4 × 1021 photons/m2 s 11.6 6.59 × 10–34 J s 11.7 (a) 3.38 × 10–19 J = 2.11 eV (b) 3.0 × 1020 photons/s 11.8 2.0 V 11.9 No, because ν < νo 11.10 4.73 × 1014 Hz 11.11 2.16 eV = 3.46 × 10–19J 11.12 (a) 4.04 × 10–24 kg m s–1 (b) 0.164 nm 11.13 (a) 5.92 × 10–24 kg m s–1 (b) 6.50 × 106 m s–1 (c) 0.112 nm 11.14 (a) 6.95 × 10–25 J = 4.34 µeV (b) 3.78 × 10–28 J = 0.236 neV 11.15 (a) 1.7 × 10–35 m (b) 1.1 × 10–32 m (c) 3.0 × 10–23 m 11.16 (a) 6.63 × 10–25 kg m/s (for both) (b) 1.24 keV (c) 1.51 eV 11.17 (a) 6.686 × 10–21 J = 4.174 × 10–2 eV (b) 0.145 nm 11.18 λ = h/p = h/(hν/c) = c/ν 11.19 0.028 nm 11.20 (a) Use eV = (m v2/2) i.e., v = [(2eV/m)]1/2 ; v = 1.33 × 107 m s–1. (b) If we use the same formula with V = 107 V, we get v = 1.88 × 109 m s–1. This is clearly wrong, since nothing can move with a speed greater than the speed of light (c = 3 × 108 m s–1). Actually, 509 the above formula for kinetic energy (m v2/2) is valid only when (v/c) << 1. At very high speeds when (v/c ) is comparable to (though always less than) 1, we come to the relativistic domain 2020-21

Physics where the following formulae are valid: Relativistic momentum p = m v Total energy E = m c 2 Kinetic energy K = m c 2 – m 0 c 2, 1 − v2  −1/ 2 c2  where the relativistic mass m is given by m = m0 m0 is called the rest mass of the particle. These relations also imply: E = (p 2c 2 + m 2 c4)1/2 0 Note that in the relativisitc domain when v/c is comparable to 1, K or energy ≥ m 0c 2 (rest mass energy). The rest mass energy of electron is about 0.51 MeV. Thus a kinetic energy of 10 MeV, being much greater than electron’s rest mass energy, implies relativistic domain. Using relativistic formulas, v (for 10 MeV kinetic energy) = 0.999 c. 11.21 (a) 22.7 cm (b) No. As explained above, a 20 MeV electron moves at relativistic speed. Consequently, the non-relativistic formula R = (m0v/e B ) is not valid. The relativistic formula is ( )R = p /eB = mv /eB or R = m0v / eB 1 – v2 /c 2 11.22 We have e V = (m v2/2) and R = (m v/e B) which gives (e/m) = (2V/R 2 B 2 ); using the given data (e/m) = 1.73 × 1011 C kg –1. 11.23 (a) 27.6 keV (b) of the order of 30 kV 11.24 Use λ = (hc/E) with E = 5.1 × 1.602 × 10–10J to get λ = 2.43 × 10–16 m. 11.25 (a) For λ = 500 m, E = (h c/ λ) = 3.98 × 10–28J. Number of photons emitted per second = 104J s–1/3.98 × 10–28J ∼ 3 × 1031 s–1 We see that the energy of a radiophoton is exceedingly small, and the number of photons emitted per second in a radio beam is enormously large. There is, therefore, negligible error involved in ignoring the existence of a minimum quantum of energy (photon) and treating the total energy of a radio wave as continuous. (b) For ν = 6 × 1014 Hz, E ∼ 4 × 10–19J. Photon flux corresponding to minimum intensity = 10–10 W m–2/4×10–19J = 2.5 × 108 m–2 s–1 Number of photons entering the pupil per second = 2.5 × 108 × 0.4 × 10–4 s–1 = 104 s–1. Though this number is not as large as in (a) above, it is large enough for us never to ‘sense’ or ‘count’ individual photons by our eye. 11.26 φ0 = h ν – e V0 = 6.7 × 10–19 J = 4.2 eV; ν0 = φ0 = 1.0 × 1015 Hz; λ = 6328Å h corresponds to ν = 4.7 × 1014 Hz < ν0. The photo-cell will not respond howsoever high be the intensity of laser light. 11.27 Use e V0 = h ν – φ0 for both sources. From the data on the first source, φ0 = 1.40 eV. Use this value to obtain for the second source V0 = 1.50 V. 510 2020-21

Answers 11.28 Obtain V0 versus ν plot. The slope of the plot is (h/e) and its intercept on the ν-axis is ν0. The first four points lie nearly on a straight line which intercepts the ν-axis at ν0 = 5.0 × 1014 Hz (threshold frequency). The fifth point corresponds to ν < ν0; there is no photoelectric emission and therefore no stopping voltage is required to stop the current. Slope of the plot is found to be 4.15 × 10–15 V s. Using e = 1.6 × 10–19 C, h = 6.64 × 10–34 J s (standard value h = 6.626 × 10–34 J s), φ0 = h ν0 = 2.11 V. 11.29 It is found that the given incident frequency ν is greater than ν0 (Na), and ν0 (K); but less than ν0 (Mo), and ν0 (Ni). Therefore, Mo and Ni will not give photoelectric emission. If the laser is brought closer, intensity of radiation increases, but this does not affect the result regarding Mo and Ni. However, photoelectric current from Na and K will increase in proportion to intensity. 11.30 Assume one conduction electron per atom. Effective atomic area ~10–20 m2 Number of electrons in 5 layers = 5×2×10−4 m2 = 1017 10−20 m2 Incident power = 10–5 W m–2 × 2 × 10–4 m2 = 2 × 10–9 W In the wave picture, incident power is uniformly absorbed by all the electrons continuously. Consequently, energy absorbed per second per electron = 2 × 10–9/1017 = 2 × 10–26 W Time required for photoelectric emission = 2 × 1.6 × 10–19J/2 × 10–26 W = 1.6 × 107 s which is about 0.5 year. Implication: Experimentally, photoelectric emission is observed nearly instantaneously (∼10–9 s): Thus, the wave picture is in gross disagreement with experiment. In the photon-picture, energy of the radiation is not continuously shared by all the electrons in the top layers. Rather, energy comes in discontinuous ‘quanta’. and absorption of energy does not take place gradually. A photon is either not absorbed, or absorbed by an electron nearly instantly. 11.31 For λ = 1 Å, electron’s energy = 150 eV; photon’s energy = 12.4 keV. Thus, for the same wavelength, a photon has much greater energy than an electron. 11.32 (a) λ=h = h Thus, for same K, λ decreases with m as p 2 mK (1/ m ). Now (mn/me) = 1838.6; therefore for the same energy, ( )(150 eV) as in Ex. 11.31, wavelength of neutron = 1/ 1838.6 × 10–10 m = 2.33×10–12 m. The interatomic spacing is about a hundred times greater. A neutron beam of 150 eV energy is therefore not suitable for diffraction experiments. 511 2020-21

Physics (b) λ = 1.45 × 10–10 m [Use λ =(h / 3 m k T ) ] which is comparable to interatomic spacing in a crystal. Clearly, from (a) and (b) above, thermal neutrons are a suitable probe for diffraction experiments; so a high energy neutron beam should be first thermalised before using it for diffraction. 11.33 λ = 5.5 × 10–12 m λ ( yellow light) = 5.9 × 10–7m Resolving Power (RP) is inversely proportional to wavelength. Thus, RP of an electron microscope is about 105 times that of an optical microscope. In practice, differences in other (geometrical) factors can change this comparison somewhat. 11.34 p = h = 6.63 ×10–34 Js = 6.63 × 10–19 kg m s–1 λ 10–15 m Use the relativistic formula for energy: E2 = c2p 2 + m 2 c4 = 9 × (6.63)2 × 10–22 + (0.511 × 1.6)2 × 10–26 0 ∼ 9 × (6.63)2 × 10–22, the second term (rest mass energy) being negligible. Therefore, E = 1.989 × 10–10 J = 1.24 BeV. Thus, electron energies from the accelerator must have been of the order of a few BeV. 11.35 Use λ = h ; mHe = 4×10 3 kg 3m kT 6 ×1023 This gives λ = 0.73 × 10–10 m. Mean separation r = (V/N)1/3 = (kT/p)1/3 For T = 300 K, p = 1.01 × 105 Pa, r = 3.4 × 10–9 m. We find r >> λ. 11.36 Using the same formula as in Exercise 11.35, λ = 6.2 × 10–9 m which is much greater than the given inter-electron separation. 11.37 (a) Quarks are thought to be confined within a proton or neutron by forces which grow stronger if one tries to pull them apart. It, therefore, seems that though fractional charges may exist in nature, observable charges are still integral multiples of e. (b) Both the basic relations e V = (1/2 ) m v 2 or e E = m a and e B v =m v 2/r, for electric and magnetic fields, respectively, show that the dynamics of electrons is determined not by e, and m separately but by the combination e/m. (c) At low pressures, ions have a chance to reach their respective electrodes and constitute a current. At ordinary pressures, ions have no chance to do so because of collisions with gas molecules and recombination. (d) Work function merely indicates the minimum energy required for the electron in the highest level of the conduction band to get out of the metal. Not all electrons in the metal belong to this level. They occupy a continuous band of levels. Consequently, for the same incident radiation, electrons knocked off from different levels come out with different energies. (e) The absolute value of energy E (but not momentum p) of any particle is arbitrary to within an additive constant. Hence, while λ is physically significant, absolute value of ν of a matter wave of an electron has no direct physical meaning. The phase speed νλ 512 2020-21

Answers is likewise not physically significant. The group speed given by dν λ ) = dE = d  p2  = p d(1/ dp dp  2m  m is physically meaningful. CHAPTER 12 12.1 (a) No different from (b) Thomson’s model; Rutherford’s model (c) Rutherford’s model (d) Thomson’s model; Rutherford’s model (e) Both the models 12.2 The nucleus of a hydrogen atom is a proton. The mass of it is 1.67 × 10–27 kg, whereas the mass of an incident α-particle is 6.64 × 10–27 kg. Because the scattering particle is more massive than the target nuclei (proton), the α-particle won’t bounce back in even in a head-on collision. It is similar to a football colliding with a tenis ball at rest. Thus, there would be no large-angle scattering. 12.3 820 nm. 12.4 5.6 × 1014 Hz 12.5 13.6 eV; –27.2 eV 12.6 9.7 × 10 – 8 m; 3.1 × 1015 Hz. 12.7 (a) 2.18 × 106 m/s; 1.09 × 106 m/s; 7.27 × 105 m/s (b) 1.52 × 10–16 s; 1.22 × 10–15 s; 4.11 × 10–15 s. 12.8 2.12×10–10 m; 4.77 × 10–10 m 12.9 Lyman series: 103 nm and 122 nm; Balmer series: 656 nm. 12.10 2.6 × 1074 12.11 (a) About the same. (b) Much less. (c) It suggests that the scattering is predominantly due to a single collision, because the chance of a single collision increases linearly with the number of target atoms, and hence linearly with thickness. (d) In Thomson’s model, a single collision causes very little deflection. The observed average scattering angle can be explained only by considering multiple scattering. So it is wrong to ignore multiple scattering in Thomson’s model. In Rutherford’s model, most of the scattering comes through a single collision and multiple scattering effects can be ignored as a first approximation. 12.12 The first orbit Bohr’s model has a radius a0 given by a0 = 4 π ε0(h /2π)2 . If we consider the atom bound by the gravitational mee2 force (Gmpme/r 2 ), we should replace (e2/4 π ε0) by Gm pm e. That is, the radius of the first Bohr orbit is given by a G = (h /2π)2 ≅ 1.2 × 1029 m. 0 Gm pm 2 e 513 2020-21

Physics This is much greater than the estimated size of the whole universe! 12.13 ν = me 4  1 − 1  = me4(2n −1) − 1)2 (4π)3ε02(h /2π)3  (n − 1)2 n2  (4π)3ε02(h /2π)3n2(n For large n, ν ≅ me 4 32 π3 ε 2 (h / 2π)3 n 3 0 Orbital frequency νc = (v/2 π r ). In Bohr model v = n(h /2π) , and mr r = 4 π ε 0(h / 2π)2 n 2 . This gives νc = n(h /2π) = 32 π3 me 4 me 2 2 πmr 2 ε02(h /2π)3n 3 which is same as ν for large n. 12.14 (a) The quantity  4 e2  has the dimensions of length. Its value  π ε 0mc 2  is 2.82 ×10–15 m – much smaller than the typical atomic size. (b) The quantity 4πε0(h /2π)2 has the dimensions of length. Its me 2 value is 0.53 × 10–10 m – of the order of atomic sizes. (Note that the dimensional arguments cannot, of course, tell us that we should use 4π and h/2π in place of h to arrive at the right size.) 12.15 In Bohr’s model, mvr = nh and mv 2 = 4 Ze 2 2 2π r π ε 0r which give T= 1 mv 2 = Ze 2 ;r = 2 8 π ε0r These relations have nothing to do with the choice of the zero of potential energy. Now, choosing the zero of potential energy at infinity we have V = – (Z e2/4 π ε 0 r) which gives V = –2T and E = T + V = – T (a) The quoted value of E = – 3.4 eV is based on the customary choice of zero of potential energy at infinity. Using E = – T, the kinetic energy of the electron in this state is + 3.4 eV. (b) Using V = – 2T, potential energy of the electron is = – 6.8 eV (c) If the zero of potential energy is chosen differently, kinetic energy does not change. Its value is + 3.4 eV independent of the choice of the zero of potential energy. The potential energy, and the total energy of the state, however, would alter if a different zero of the potential energy is chosen. 514 12.16 Angular momenta associated with planetary motion are incomparably large relative to h . For example, angular momentum of the earth in its orbital motion is of the order of 1070h . In terms of the Bohr’s quantisation postulate, this corresponds to a very large value of n (of the order of 1070). For such large values of n, the differences in the successive energies and angular momenta of the quantised levels of the Bohr model are so small compared to the energies and angular momenta respectively for the levels that one can, for all practical purposes, consider the levels continuous. 2020-21

Answers 12.17 All that is needed is to replace m e by mµ in the formulas of the Bohr model. We note that keeping other factors fixed, r ∝ (1/m) and E ∝ m. Therefore, = re me 0.53 ×10 13 = 2.56 × 10–13 m mµ 207 rµ = Eµ= Ee m µ = – (13.6 × 207) eV ≅ – 2.8 keV me CHAPTER 13 13.1 (a) 6.941 u (b) 19.9%, 80.1% 13.2 20.18 u 13.3 104.7 MeV 13.4 8.79 MeV, 7.84 MeV 13.5 1.584 × 1025 MeV or 2.535×1012J 13.6 i) 226 Ra → 222 Rn + 4 He ii) 242 Pu → U238 + 4 He 88 86 2 94 92 2 iii) 32 P → 1362S + e– + ν iv) 210 B → 210 Po + e– +ν 15 83 84 v) 161C → 151B + e+ +ν vi) 97 Tc → 97 Mo + e+ +ν 43 42 vii) 120 Xe + e – → I120 + ν 54 53 13.7 (a) 5 T years (b) 6.65 T years 13.8 13.9 4224 years 13.10 13.11 7.126 ×10–6 g 13.12 7.877 ×1010 Bq or 2.13 Ci 13.13 1.23 (a) Q = 4.93 MeV, Eα = 4.85 MeV (b) Q = 6.41 MeV, Eα = 6.29 MeV 11 C → 151B + e+ +ν +Q 6 , where the masses used are those of nuclei and not of atoms. If we 161C use atomic masses, we have to add 6me in case of and 5m e in case of 115B. Hence (Note me has been doubled) Using given masses, Q = 0.961 MeV. Q = Ed+ Ee+ Eν The daughter nucleus is too heavy compared to e+ and ν, so it carries negligible isenmerignyim(Eudm≈0(i)..eI.f, the kinetic energy (Ecaν)rrciaersrimedaxbiymuthme neutrino zero), the positron energy, and this is practically all energy Q; hence maximum Ee ≈ Q). ( ) ( )13.14 23 23 – me  c2 , where the 23 Ne → 23 Na + e– +ν +Q ; Q = m N 10 Ne – mN 11 Na 10 11 masses used are masses of nuclei and not of atoms as in Exercise ( ) ( )13.13. 23 Ne 23 Na  c 2 . Using atomic masses Q = m 10 –m 11 Note me has 515 2020-21

Physics been cancelled. Using given masses, Q = 4.37 MeV. As in Exercise 13.13, maximum kinetic energy of the electron (max Ee) = Q = 4.37 MeV. 13.15 (i) Q = – 4.03 MeV; endothermic (ii) Q = 4.62 MeV; exothermic ( ) ( )13.16 Q = 56 28 m 26 Fe – 2m 13 Al = 26.90 MeV; not possible. 13.17 4.536 × 1026 MeV 13.18 Energy generated per gram of U235 = 6 × 1023 × 200 × 1.6 × 10−13 J g−1 235 92 The amount of U235 consumed in 5y with 80% on-time 92 = 5 × 0.8 × 3.154 × 1016 × 235 g = 1544 kg 1.2 × 1.6 × 1013 The initial amount of U235 = 3088 kg. 92 13.19 About 4.9 × 104 y 13.20 360 KeV 13.22 Consider the competing processes: A X → A –1 Y + e+ + νe + Q1 (positron emission) Z Z e– + A X → A –1 Y + νe + Q2 (electron capture) Z Z ( ) ( )Q1 – me  c2 = m N A X – mN A –1 Y Z Z ( ) ( )= m A X –m A –1 Y – 2me  c 2 Z Z ( ) ( ) ( ) ( )Q2  c2  c2 = m N A X + me – mN A –1 Y = m A X –m A –1 Y Z Z Z Z This means Q1 > 0 implies Q2 > 0 but Q2 > 0 does not necessarily mean Q1 > 0. Hence the result. 13.23 25 Mg : 9.3%, 26 Mg :11.7% 12 12 13.24 Neutron separation energy Sn of a nucleus A X is Z Sn = m N ( A −1 X) + m n − mN( A X) c 2 Z Z From given data , Sn ( 2401Ca) = 8.36MeV,Sn ( 27 Al) = 13.06MeV 13 13.25 209 d 13.26 For 14 C emission 6 Q = [m ( 223 Ra ) − m ( 209 Pb) − m ( 14 C)]c 2 88 82 6 N N N = [m( 223 Ra ) − m( 209 Pb) − m ( 14 C)]c 2 = 31.85 MeV 88 82 6 For 4 He emission, Q = [m ( 223 Ra) − m( 219 Rn) − m( 4 He)]c 2 = 5.98MeV 2 88 86 2 13.27 Q = [m ( 238 U ) + m − m ( 140 Ce) − m ( 99 Ru)]c 2 = 231.1 MeV 92 58 44 n 516 2020-21

Answers 13.28 (a) Q = [m ( 2 H) + m ( 31H) − m ( 4 He) − m n ]c 2 = 17.59 MeV (b) 1 2 K.E. required to overcome Coulomb repulsion = 480.0 keV 480.0 KeV = 7.68×10–14 J = 3kT ∴ T = 3 7.68 ×10–14 (as k = 1.381 × 10–23 J K –1) × 1.381 × 10 –23 = 1.85 ×109 K (required temperature) ( ) ( )13.29 Kmax β1– = 0.284 MeV, Kmax β2– = 0.960 MeV ν (γ1) = 2.627 × 1020 Hz , ν (γ 2 ) = 0.995 × 1020 Hz , ν (γ 3 ) = 1.632 × 1020 Hz 13.30 (a) Note that in the interior of Sun, four 11H nuclei combine to form one 4 He nucleus releasing about 26 MeV of energy per event. 2 Energy released in fusion of 1kg of hydrogen = 39 ×1026 MeV (b) Energy released in fission of 1kg of 235 U = 5.1×1026 MeV 92 The energy released in fusion of 1 kg of hydrogen is about 8 times that of the energy released in the fission of 1 kg of uranium. 13.31 3.076 × 104 kg CHAPTER 14 14.1 (c) 14.2 (d) 14.3 (c) 14.4 (c) 14.5 (c) 14.6 50 Hz for half-wave, 100 Hz for full-wave 14.7 No (hν has to be greater than Eg ). 14.8 ne ≈ 4.95 × 1022; nh = 4.75 × 109 ; n-type since ne >> nh For charge neutrality ND – NA = ne – nh ; ne.nh = ni2 Solving these equations, ne = 1 (N D − NA )+ (N D − NA )2 + 4ni2  2  14.9 About 1 × 105 14.10 (a) 0.0629 A, (b) 2.97 A, (c) 0.336 Ω (d) For both the voltages, the current I will be almost equal to I0, showing almost infinite dynamic resistance in the reverse bias. 14.12 NOT ; AY 01 14.13 (a) AND 10 14.14 OR gate 14.15 (a) NOT, (b) OR (b) AND 517 2020-21

Physics BIBLIOGRAPHY TEXTBOOKS For additional reading on the topics covered in this book, you may like to consult one or more of the following books. Some of these books however are more advanced and contain many more topics than this book. 1 Ordinary Level Physics, A.F. Abbott, Arnold-Heinemann (1984). 2 Advanced Level Physics, M. Nelkon and P. Parker, 6th Edition, Arnold-Heinemann (1987). 3 Advanced Physics, Tom Duncan, John Murray (2000). 4 Fundamentals of Physics, David Halliday, Robert Resnick and Jearl Walker, 7th Edition John Wily (2004). 5 University Physics (Sears and Zemansky’s), H.D. Young and R.A. Freedman, 11th Edition, Addison—Wesley (2004). 6 Problems in Elementary Physics, B. Bukhovtsa, V. Krivchenkov, G. Myakishev and V. Shalnov, MIR Publishers, (1971). 7 Lectures on Physics (3 volumes), R.P. Feynman, Addision – Wesley (1965). 8 Berkeley Physics Course (5 volumes) McGraw Hill (1965). a. Vol. 1 – Mechanics: (Kittel, Knight and Ruderman) b. Vol. 2 – Electricity and Magnetism (E.M. Purcell) c. Vol. 3 – Waves and Oscillations (Frank S. Crawford) d. Vol. 4 – Quantum Physics (Wichmann) e. Vol. 5 – Statistical Physics (F. Reif ) 9 Fundamental University Physics, M. Alonso and E. J. Finn, Addison – Wesley (1967). 10 College Physics, R.L. Weber, K.V. Manning, M.W. White and G.A. Weygand, Tata McGraw Hill (1977). 11 Physics: Foundations and Frontiers, G. Gamow and J.M. Cleveland, Tata McGraw Hill (1978). 12 Physics for the Inquiring Mind, E.M. Rogers, Princeton University Press (1960). 13 PSSC Physics Course, DC Heath and Co. (1965) Indian Edition, NCERT (1967). 14 Physics Advanced Level, Jim Breithampt, Stanley Thornes Publishers (2000). 15 Physics, Patrick Fullick, Heinemann (2000). 16 Conceptual Physics, Paul G. Hewitt, Addision—Wesley (1998). 17 College Physics, Raymond A. Serway and Jerry S. Faughn, Harcourt Brace and Co. (1999). 18 University Physics, Harris Benson, John Wiley (1996). 19 University Physics, William P. Crummet and Arthur B. Western, Wm.C. Brown (1994). 20 General Physics, Morton M. Sternheim and Joseph W. Kane, John Wiley (1988). 51281 Physics, Hans C. Ohanian, W.W. Norton (1989). 2020-21

BibAlinosgwraeprshy 22 Advanced Physics, Keith Gibbs, Cambridge University Press (1996). 23 Understanding Basic Mechanics, F. Reif, John Wiley (1995). 24 College Physics, Jerry D. Wilson and Anthony J. Buffa, Prentice Hall (1997). 25 Senior Physics, Part – I, I.K. Kikoin and A.K. Kikoin, MIR Publishers (1987). 26 Senior Physics, Part – II, B. Bekhovtsev, MIR Publishers (1988). 27 Understanding Physics, K. Cummings, Patrick J. Cooney, Priscilla W. Laws and Edward F. Redish, John Wiley (2005). 28 Essentials of Physics, John D. Cutnell and Kenneth W. Johnson, John Wiley (2005). GENERAL BOOKS For instructive and entertaining general reading on science, you may like to read some of the following books. Remember however, that many of these books are written at a level far beyond the level of the present book. 1 Mr. Tompkins in paperback, G. Gamow, Cambridge University Press (1967). 2 The Universe and Dr. Einstein, C. Barnett, Time Inc. New York (1962). 3 Thirty years that Shook Physics, G. Gamow, Double Day, New York (1966). 4 Surely You’re Joking, Mr. Feynman, R.P. Feynman, Bantam books (1986). 5 One, Two, Three… Infinity, G. Gamow, Viking Inc. (1961). 6 The Meaning of Relativity, A. Einstein, (Indian Edition) Oxford and IBH Pub. Co. (1965). 7 Atomic Theory and the Description of Nature, Niels Bohr, Cambridge (1934). 8 The Physical Principles of Quantum Theory, W. Heisenberg, University of Chicago Press (1930). 9 The Physics—Astronomy Frontier, F. Hoyle and J.V. Narlikar, W.H. Freeman (1980). 10 The Flying Circus of Physics with Answer, J. Walker, John Wiley and Sons (1977). 11 Physics for Everyone (series), L.D. Landau and A.I. Kitaigorodski, MIR Publisher (1978). Book 1: Physical Bodies Book 2: Molecules Book 3: Electrons Book 4: Photons and Nuclei. 12 Physics can be Fun, Y. Perelman, MIR Publishers (1986). 13 Power of Ten, Philip Morrison and Eames, W.H. Freeman (1985). 14 Physics in your Kitchen Lab., I.K. Kikoin, MIR Publishers (1985). 15 How Things Work: The Physics of Everyday Life, Louis A. Bloomfield, John Wiley (2005). 16 Physics Matters: An Introduction to Conceptual Physics, James Trefil and Robert M. Hazen, John Wiley (2004). 519 2020-21

Physics INDEX Absorption spectra 421 Biot-Savart law 143 AC current 233 Bohr magneton 163 AC Generator 224 Bohr radius 425 AC voltage 233 Bohr’s model of atom 422 241 Bohr’s postulates 424 applied to a capacitor 234 Brewster’s angle 380 applied to a resistor 237 Brewster’s law 381 applied to an inductor 244 C.A. Volta applied to a series LCR circuit 142 Capacitance 53 Accelerators in India 447 Capacitive reactance 73 Activity of radioactive substances 241 Additivity of charges 8 Capacitive circuit 252 Alpha decay 449 Capacitor Alpha particle scattering 415 parallel plate 74 Ammeter 165 in parallel 79 Ampere 155 in series 78 Amperes circuital law 147 Cartesian sign convention 311 Analog signal 491 Cassegrain telescope 341 AND gate 492 Cells 110 Andre, Ampere 148 in parallel 114 Angle 113 of deviation in series 453 of incidence 330 Chain reaction of reflection 355 Charging by induction 6 of refraction 357 Charles August de Coulomb 11 Angular magnification 355 Chromatic aberration 332 Apparent depth 339 Coercivity 195 Area element vector 318 Coherent source 360 Atomic 103 mass unit 26 Colour code of resistors 328 number Combination of lenses spectra 107 Aurora Borealis 439 Combination of resistors 108 Band gap 440 series 438 Bar magnet 420 parallel 312 as solenoid 139 Composition of nucleus 469 Barrier potential 471 Concave mirror 97, 468 Becquerel 174 Conduction band Beta decay 176 Conductivity 5 Binding energy per nucleon 479 Conductors 8 448 Conservation of charge 51 520 450 Conservative force 32 444 Continuous charge distribution 2020-21