1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 1.10 1.11 1.12 1.13 1.14 1.15 1.16 1.17 ANSWERS Chapter 1 (a) (a) (d) (b) (c) (a) (a) (c), (d) (b), (d) (b), (d) (c), (d) (a), (c). (a), (b), (c) and (d). Zero. –QQ (i) 2 (ii)4π R14π R22 The electric fields bind the atoms to neutral entity. Fields are caused by excess charges. There can be no excess charge on the inter surface of an isolated conductor. No, the field may be normal. However, the converse is true. 106 1.18 Top view Side view qq q q1.19 (i) (ii) (iii) (iv) .8ε04ε02ε02ε0 1.20 1 Molar mass M of Al has NA = 6.023 × 1023 atoms. m ∴ m = mass of Al paisa coin has N =N atomsA M Now, ZAl = 13, MAl = 26.9815g 0.75 Hence N = 6.02 × 1023 atoms/mol × 26.9815g/mol = 1.6733 × 1022 atoms ∴ q = +ve charge in paisa = N Ze 22–19= (1.67 × 10)(13) (1.60 × 10C) = 3.48 × 104 C. q = 34.8 kC of ±ve charge. This is an enormous amount of charge. Thus we see that ordinary neutral matter contains enormous amount of ± charges. 42q 2 ⎛ 9 Nm ⎞ (3.48 ×10C) 23 1.21 (i) F1 == ⎜ 8.99 ×10 2 ⎟ = 1.1 10 × N2 –4 24πε 0 r1 ⎝ C ⎠ 10 m 2 –2 2F2 r1 (10 m) –8 –8 15 (ii) 2 2210 ⇒ F2 = F1 ×10 =× N== = 1.1 10 F1 r2 (10 m) 2 –2 2F3 r1 (10 m) –16 (iii) = 2 = 62 = 10 F1 r3 (10 m) F3 = 10–16 F1 = 1.1 × 107 N. Conclusion: When separated as point charges these charges exert an enormous force. It is not easy to disturb electrical neutrality. 1.22 (i) Zero, from symmetry. (ii) Removing a +ve Cs ion is equivalent to adding singly charged -ve Cs ion at that location. Net force then is 2 e F = 4πε 0r2 where r = distance between the Cl ion and a Cs ion. = 0.346 × 10–9 m Hence, F 9 –19 2 –9 2 (8.99×10 )(1.6 10 ) (0.346×10 ) × = –11 192 10 = × = 1.92 × 10–9 N Ans 1.92 × 10–9 N, directed from A to Cl – 1.23 At P: on 2q, Force due to q is to the left and that due to –3q is to the right. ∴ 2 2 0 2 4 q xπε 2 2 0 6 4 ( ) q d xπε = + ∴ (d + x)2 = 3x2 ∴ 2x2 – 2dx – d2 = 0 P 2q q –3q x d 2 = ± d3 2 (–ve sign would be between q and –3q and hence is unaceptable.) x d 2 = + d3 2 d (1 2 = + 3) to the left of q. 1.24 (a) Charges A and C are positive since lines of force emanate from them. (b) Charge C has the largest magnitude since maximum number of field lines are associated with it. (c) (i) near A. There is no neutral point between a positive and a negative charge. A neutral point may exist between two like charges. From the figure we see that a neutral point exists between charges A and C. Also between two like charges the neutral point is closer to the charge with smaller magnitude. Thus, electric field is zero near charge A. 1.25 (a) (i) zero (ii) 2 0 1 q alongOA 4 rπε uuuur (iii) 2 0 1 2q alongOA 4 rπε uuuur (b) same as (a). 1.26 (a) Let the Universe have a radius R. Assume that the hydrogen atoms are uniformly distributed. The charge on each hydrogen atom is eH = – (1 + y) e + e = – ye = |ye| The mass of each hydrogen atom is ~ mp (mass of proton). Expansion starts if the Coulumb repulsion on a hydrogen atom, at R, is larger than the gravitational attraction. Let the Electric Field at R be E. Then 4 π R 3N ye4πR2 E = (Gauss’s law)3εo 1N ye E (R) = R rˆ 3 εo Let the gravitational field at R be GR. Then 3 – 4πR2 G R = 4 πG mp 4 π RN 3 4G = – πGm NRR3 ρ 4 GR(R)= – πGmρNR rˆ 3 Thus the Coulombic force on a hydrogen atom at R is 221Ny e yeE(R) = R rˆ 3 εo The gravitional force on this atom is 4π 2m G ( )= – GNm R rˆ pRR p3 The net force on the atom is 22⎛ 1Nye 4π 2 ⎞ = ⎜ p ⎟F R– GNmR rˆ 3 ε 3⎝ o ⎠ The critical value is when 1 Ny 2Ce2 4π 2R = GNm R 3 εo3 p 2m ⇒ yc2 = 4πεoG 2p e –11 26 –627 ×10 ×1.8 ×10 × 81 10× 9 2 –389 ×10 ×1.6 ×10 � 63 ×10–38 –19 –18∴ yC � 8 ×10 � 10 (b) Because of the net force, the hydrogen atom experiences an acceleration such that 2 22dR ⎛ 1Nye 4p ⎞ m = R– GNm 2R p2 ⎜ p ⎟ dt 3e 3⎝ o ⎠ 2 22 22 2dR 1 ⎛ 1Nye 4p ⎞ Or, 2 =aR where α = ⎜ – GNm p ⎟dt m3e3 p ⎝ o ⎠ at –at This has a solution R=Ae +Be As we are seeking an expansion, B = 0. t∴ R = Aeα . ⇒ R =α Ae αt =αR Thus, the velocity is proportional to the distance from the centre. 1.27 (a) The symmetry of the problem suggests that the electric field is radial. For points r < R, consider a spherical Gaussian surfaces. Then on the surface E .dS =ρdv �∫ 1 ∫Vr ε o r 21 34πr Er = 4πk∫r′ dr ′ ε o o 14 πk 4 = r ε 4 o 12∴ E = kr r 4ε o 12E ( ) = kr rˆr 4ε o For points r > R, consider a spherical Gaussian surfaces’ of radius r, 1 E .dS = ρdv �∫ r ∫ε oV R 24πk 34πr Er = ∫r dr ε o o 4πkR4 = ε o 4 4kR ∴ E = r 4ε or 2 E () (/4 42ˆr = k ε o )( R /r )r y1.28 d -Q qQ (b) The two protons must be on the opposite sides of the centre along a diameter. Suppose the protons are at a distance r from the centre. R Now, 4π kr ′3dr = 2e∫ o 4πk ∴ R4 = 2e 4 2e ∴k = 4πR Consider the forces on proton 1. The attractive force due to the charge distribution is e 22e 2 r 2 –e E = – kr rˆ = – rˆ r 4ε 4πε R4 oo e 21 rˆThe repulsive force is 4πε o (2r )2 2 22⎛ e 2er ⎞ – rˆNet force is ⎜ 24 ⎟4πε 4r 4πε R⎝ oo ⎠ This is zero such that 2 22 e 2er = 16 πε r 24πε R4 oo 44R4 R4 Or, r == 32 8 R⇒ r = (8)1/4 RThus, the protons must be at a distance r =from the centre. 48 Q(a) The electric field at γ due to plate α is – xˆ S2ε o The electric field at γ due to plate β is q xˆ S2ε o Hence, the net electric field is (Q – q)E1 =(–xˆ )2ε S o (b) During the collision plates β & γ are together and hence must be at one potential. Suppose the charge on β is q1 and on γ is q2 . Consider a point O. The electric field here must be zero. QElectric field at 0 due to α= – xˆ 2ε S o q1ˆElectric field at 0 due to β= – x 2ε oS y qElectric Field at 0 due to γ= –2 xˆ °o2ε oS q1 q2 – (Q + q2 ) q∴+ 1 = 0 2ε S 2ε S oo ⇒ q1– q2 = Q Further, q1 + q2 = Q + q ⇒ q1 = Q + q /2 and q2 = q/2 Thus the charge on β and γ are Q + q/2 and q/2, respectively. (c) Let the velocity be v at the distance d after the collision. If m is the mass of the plate γ, then the gain in K.E. over the round trip must be equal to the work done by the electric field. After the collision, the electric field at γ is Q (Q + q /2 ) q /2 E2 = – xˆ + xˆ = xˆ 2ε S 2ε S 2ε S oo o The work done when the plate γ is released till the collision is F1d where F1 is the force on plate γ. The work done after the collision till it reaches d is F2d where F2 is the force on plate γ . (Q – q) Q F = EQ = 112ε S o )2(q /2 and F = Eq /2 = 22 2ε S o ∴ Total work done is 1 21⎡(Q – q)Q +(q /2 ) ⎤ d =(Q – q /2 )2 d⎣⎦2ε oS 2ε oS d2⇒ (1/2) mv =(Q – q /2 )2 2ε S o 1/2 ⎛ d ⎞ ∴v =(Q – q /2 )⎜⎟ mε S⎝ o ⎠ Qq [1esuof charge] 2 1.29 (i) F = 2 = 1dyne = 2 r [1cm] Or, 1 esu of charge = 1 (dyne)1/2 (cm) Hence, [1 esu of charge] = [F]1/2 L = [MLT–2]1/2 L = M1/2 L3/2 T–1 [1 esu of charge] = M1/2 L3/2 T–1 Thus charge in cgs unit is expressed as fractional powers (1/2) of M and (3/2) of L. (ii) Consider the coloumb force on two charges, each of magnitude 1 esu of charge separated by a distance of 1 cm: The force is then 1 dyne = 10–5 N. This situation is equivalent to two charges of magnitude x C separated by 10–2m. This gives: 2 –4 0 1 . 4 10 x F πε = which should be 1 dyne = 10–5 N. Thus x 2 –4 0 1 . 4 10 πε –5 0 110 4πε = ⇒ x –9 2 2 2 10 Nm C = With x 9 1 [3] 10 = × , this yields 2 –9 2 18 2 9 2 0 1 Nm 10 [3] 10 [3] 10 4 Cπε = × × = × With [3] → 2.99792458, we get 2 9 2 0 1 Nm 8.98755.... 10 4 Cπε = × exactly 1.30 Net force F on q towards the centre O F q r 2 2 0 2 cos 4 θ πε = q x r r 2 2 0 2 – . 4πε = q – q d x d – q F = 2 2 2 3/2 0 –2 4 ( ) q x d xπε + q x kx d 2 3 0 –2 – 4πε ≈ = for x << d. Thus, the force on the third charge q is proportional to the displacement and is towards the centre of the two other charges. Therefore, the motion of the third charge is harmonic with frequency 2q 2 k ω= = 34πε 0dm m 1/2 332π ⎡8πε md ⎤and hence 0 .T = ⎢ 2 ⎥ω q⎣⎦ 1.31 (a) Slight push on q along the axis of the ring gives rise to the situation shown in Fig (b). A and B are two points on the ring at the end of a diameter. –QForce on q due to line elements at A and B is Z Axis of the ring Total charge –Q 2π R–Q 11 F = 2. .. q . .cos θA+B 22π R 4πε 0 r –Q q 1 z = ..2 2 2 21/2 π R.4 πε 0 (z + R )( z + R ) Total force due to ring on q = (FA+B)(πR) –Qq z = 4πε0 2 23/2 (a) (z + R ) –Qq for z << R 4πε 0Thus, the force is propotional to negative of displacement. Motion under such forces is harmonic. (b) From (a) d2 z Qqz d2 z Qq m = – or = – z232 3dt 4πε 0R dt 4πε 0mR Qq 4πε mR That is, 2 . Hence 0ω= 3 T = 2π 4πε 0mR Qq (b) Chapter 2 2.1 (d) 2.2 (c) 2.3 (c) 2.4 (c) 2.5 (a) 2.6 (c) 2.7 (b), (c), (d) 2.8 (a), (b), (c) 2.9 (b), (c) 2.10 (b), (c) 2.11 (a), (d) 2.12 (a), (b) 2.13 (c) and (d) 2.14 More. 2.15 Higher potential. 2.16 Yes, if the sizes are different. 2.17 No. 2.18 As electric field is conservative, work done will be zero in both the cases. 2.19 Suppose this were not true. The potential just inside the surface would be different from that at the surface resulting in a potential gradient. This would mean that there are field lines pointing inwards or outwards from the surface. These lines cannot at the other end be again on the surface, since the surface is equipotential. Thus, this is possible only if the other end of the lines are at charges inside, contradicting the premise. Hence, the entire volume inside must be at the same potential. 2.20 C will decrease 1 2Energy stored = CV and hence will increase.2 Electric field will increase. Charge stored will remain the same. U V will increase. 2.21 Consider any path from the charged conductor to the uncharged conductor along the electric field. The potential will continually decrease along this path. A second path from the uncharged conductor to infinity z will again continually lower the potential further. Hence this result. −qQ2.22 U= 2 24πε R + z R0 z The variation of potential energy with z is shown in the figure. P The charge – q displaced would perform oscillations. We cannot conclude r Q anything just by looking at the graph. 1 Q2.23 V = 4πε 22 0 R + z 2.24 To find the potential at distance r from the line consider the electric field. We note that from symmetry the field lines must be radially outward. Draw a cylindrical Gaussian surface of radius r and length l. Then 1∫ E.d S =λl ε0 1Or E2 π rl =λlr ε0 λ⇒ Er = 2πε 0r Hence, if r0 is the radius, r λ r V (r) – V(r ) =− ∫ E.d l = ln 0 02πε r r00 For a given V, r 2πε 0ln =− [V(r) – V(r )] 0 0r λ −2πε Vr /λ+2πε V(r) /λ0 0 0⇒ r = r0e .e The equipotential surfaces are cylinders of radius −2πε 0[V(r)–V(r 0)]/ λ r = r e0 2.25 Let the plane be at a distance x from the origin. The potential at the point P is 1 q 1 q−1/2 1/2 4πε 22 4πε 220 ⎡⎣(x + d /2 )+ h ⎤⎦ 0 ⎡⎣(x − d /2 )+ h ⎤⎦ If this is to be zero. 11 = 1/2 1/2 22 22⎡(x + d /2 )+ h ⎤⎡(x − d /2 )+ h ⎤⎣ ⎦⎣ ⎦ Or, (x-d/2)2 + h2 = (x+d/2)2 + h2 222 2⇒ x − dx + d /4 = x + dx + d /4 Or, 2dx = 0 ⇒ x = 0 The equation is that of a plane x = 0. 2.26 Let the final voltage be U: If C is the capacitance of the capacitor without the dielectric, then the charge on the capacitor is Q1 = CU The capacitor with the dielectric has a capacitance εC. Hence the charge on the capacitor is Q2 =ε U =α CU 2 The initial charge on the capacitor that was charged is Q0 = CU0 From the conservation of charges, Q0 = Q1 + Q2 Or, CU0 = CU + α CU2 ⇒α U 2 + U − u0 = 0 −1± 1+ 4αU0∴U = 2α = volts 4 As U is positive 625 −1 24 U = == 6V 4 4 2.27 When the disc is in touch with the bottom plate, the entire plate is a equipotential. A change q′ is transferred to the disc. The electric field on the disc is V = d ∴ q′ = − ε V πr 2 0 d The force acting on the disc is 2VV −× q′=ε02 πr 2 dd If the disc is to be lifted, then 2V επr 2 = mg 02d mgd 2 ⇒ V = 2πε r0 1 qq qq ⎫⎧ dd qq ud ud2.28 U= ⎨ –– ⎬4πε rrr ⎭0 ⎩ 9×10 9 –19 = 10 –15 (1.6×10 )2 {(1 3) 2 –(2 3)(1 3)–(2 3)(1 3) } ⎧14⎫ –14 = 2.304 × 10–13 ⎨ – ⎬=–7.68×10 J ⎩99⎭ = 4.8 × 105 eV = 0.48 MeV = 5.11 × 10–4 (mnc2) 2.29 Before contact Q1=σ .4 π R2 Q2=σ .4 π (2 R2) = 4(σ .4 π R2 )= 4Q1 After contact : ′ ,Q ′+ Q = Q + Q =5Q 1212 1 =5 (σ .4 π R2) They will be at equal potentials: ′′Q1 Q2 = R 2R ∴Q2 ′=2Q ′ . 3Q1 ′=5(π 2 )∴σ.4 R′ 5 210 2(σ.4 R ) Q′= σ.4 R )∴ Q =π and (π13 23 5 ∴σ1 =53 σ and ∴σ 2 =σ .6 12.30 Initially : V ∝ and V1 + V2 = E C ⇒V1=3V and V2 =6V ∴ Q =CV = 6C×3 =18 µ C C11 11 K1 K2 Q2 = 9 µ C and Q 3 = 0 Later : Q =Q′+ Q E = 9V C3 C2223 Q2with C2V + C3V = Q2 ⇒V = = (3 2)V C +C Q2 ′= (9 2) µC and Q 3 ′= (9 2) µC 23 Q2.31 σ= 2π R 1 σ.2 πr dr dU = 4 πε 0 r 2 + z 2 πσ R2rdr ∴U = 4πε ∫ 22 0O r + z z q1 d O q2 2.32–d 2q 2 U = 04πε 0d 2πσ R2πσ ⎡ 22 ⎤⎡ 22 ⎤= r + z = R +z –z⎥⎦ 2Q ⎡ 22 ⎤ 4πε 0 ⎣⎢ ⎦⎥O4πε 0 ⎣⎢ = R +z –z2 ⎢ ⎥⎦4πε 0R ⎣ qq12 = 0+ 22 22 x +y +(z – d )2 x + y +(z + d )2 q –q12∴= 22 22 x + y +(z – d )2 x + y +(z + d )2 Thus, to have total potential zero, q1 and q2 must have opposite signs. Squaring and simplifying, we get. 2 ⎤⎡(qq2 )+12222 1 x + y + z + ⎢⎥(2 zd )+d = 0 ⎢(qq2 )2 –1⎥⎦⎣ 1 ⎛⎡q12 + q12 ⎤⎞ This is the equation of a sphere with centre at ⎜⎜0,0,–2 d ⎢ 22 ⎥ ⎟⎟ . q – q⎝⎣ 11 ⎦⎠ Note : if q1 = –q2 ⇒ Then z = 0, which is a plane through mid-point. 1 ⎧⎪ –q 2–q 2 ⎫⎪U = ⎨ + ⎬4πε 0 ⎪⎩(d – x ) (d – x ) ⎪⎭ –q 22dU = 4πε 0 (d2– x 2 ) dU –q 2.2 d 2x = . dx 4π∈ 2 22 0 (d – x ) dU =0 at x = 0 dx x = 0 is an equilibrium point. 22 ⎡ 28x 2 ⎤dU ⎛ –2 dq ⎞ – = ⎢ 2 223 ⎥2 ⎜⎟ 2 2(d – x )dx ⎝ 4π∈0 ⎢(d – x ) ⎥⎠⎣ ⎦ ⎛ –2 dq 2 ⎞ 12⎡ 22 2 ⎤= ⎜⎟ 2 23 2(d – x ) –8x ⎝ 4π∈0 ⎠ (d – x ) ⎣ ⎦ At x = 0 d2U ⎛ –2 dq 2 ⎞⎛ 1 ⎞ 2 =⎜ ⎟⎜ (2 d )26 ⎟ , which is < 0.dx 4π∈ ⎝ d ⎠⎝ 0 ⎠ Hence, unstable equilibrium. Chapter 3 3.1 (b) 3.2 (a) 3.3 (c) 3.4 (b) 3.5 (a) 3.6 (a) 3.7 (b), (d) 3.8 (a), (d) 3.9 (a), (b) 3.10 (b), (c) 3.11 (a), (c) 3.12 When an electron approaches a junction, in addition to the uniform E that it normally faces (which keep the drift velocity v fixed), thered are accumulation of charges on the surface of wires at the junction. These produce electric field. These fields alter direction of momentum. 3.13 Relaxation time is bound to depend on velocities of electrons and ions. Applied electric field affects the velocities of electrons by speeds at the order of 1mm/s, an insignificant effect. Change in T, on the other hand, affects velocities at the order of 102 m/s. This can affect τ significantly. [ρ = ρ(E,T ) in which E dependence is ignorable for ordinary applied voltages.] 3.14 The advantage of null point method in a Wheatstone bridge is that the resistance of galvanometer does not affect the balance point and there is no need to determine current in resistances and galvanometer and the internal resistance of a galvanometer. R can be calculatedunknown120 applying Kirchhoff ’s rules to the circuit. We would need additional accurate measurement of all the currents in resistances and galvanometer and internal resistance of the galvanometer. 3.15 The metal strips have low resistance and need not be counted in the potentiometer length l1 of the null point. One measures only their lengths along the straight segments (of lengths 1 meter each). This is easily done with the help of centimeter rulings or meter ruler and leads to accurate measurements. 3.16 Two considerations are required: (i) cost of metal, and (ii) good conductivity of metal. Cost factor inhibits silver. Cu and Al are the next best conductors. 3.17 Alloys have low value of temperature co-efficient (less temperature sensitivity) of resistance and high resistivity. 3.18 Power wasted PC = I2RC where RC is the resistance of the connecting wires. 2P P = RC 2 C V In order to reduce PC, power should be transmitted at high voltage. 3.19 If R is increased, the current through the wire will decrease and hence the potential gradient will also decrease, which will result in increase in balance length. So J will shift towards B. 3.20 (i) Positive terminal of E1 is connected at X and E1 > E. (ii) Negative terminal of E1 is connected at X. V 3.21 E R E E 3.22 I = ; = 10 I RR + nR R + n 1+ n 1+ n = 10 = n = n1 n +11 + n ∴ n = 10. 11 1 RRR Rmin min min min 3.23 =+ ....... + , = ++ ....... +> 1 RR RRRR R p 1 nP 12 n and R = R + ...... + R ≥ R .S1nmax In Fig. (b), Rmin provides an equivalent route as in Fig. (a) for current. But in addition there are (n – 1) routes by the remaining (n – 1) resistors. Current in Fig.(b) > current in Fig. (a). Effective Resistance in Fig. (b) < Rmin. Second circuit evidently affords a greater resistance. You can use Fig. (c) and (d) and prove R s > R max. RRRmin maxmax Rmin V VV (a) (b) (c) (d) 3.24 6 – 4 0.2A 2 8 I = = + A B P.D. across E 1 = 6 – 0.2 × 2 = 5.6 V P.D. across E 2 = VAB = 4 + 0.2 × 8 = 5.6 V Point B is at a higher potential than A E1 E2 3.25 I = 1 2 E E R r r + + + V 1 = E – Ir1 = 1 1 2 2E – 0 R E r r r = + + or E 1 1 2 2E R r r r = + + r r 1 1 2 2r 1 R = + + r 1 + r 2 + R = 2r 1 R = r1 – r2 3.26 AR l –3 2(10 0.5) ρ π = × BR l –3 2 –3 2 [(10 ) – (0.5 10 ) ] ρ π = × A B R R –3 2 –3 2 –3 2 (10 ) – (0.5 10 ) (.5 10 ) × = × = 3 : 1 3.27 R I Veff Reff 3.28 3.29 We can think of reducing entire network to a simple one for any branch R as shown in Fig. V Then current through R is I = eff Reff + R Dimensionally V = V (V, V, ...... V) has a dimension of voltage andeffeff12nR = R (R , R , ....... R ) has a dimension of resistance. effeff12mTherefore if all are increased n-fold new new V = nV ,R = n Reff eff eff eff and Rnew = nR. Current thus remains the same. Applying Kirchhoff’s junction rule: I = I + I 12 Kirchhoff’s loop rule gives: 10 = IR + 10I1....(i) 2 = 5I2 – RI = 5 (I1 – I) – RI 4 = 10I1 – 10I – 2RI..... (ii) ⎛ 10 ⎞(i) – (ii) ⇒ 6 = 3RI + 10I or,2 = I ⎜ R+ ⎟⎝ 3 ⎠ 2 = (R+ R)I Comparing with V = (R + R )Ieffeff effand Veff = 2V 10 Reff = Ω . 3 R R 2V I1 I I2 I2 10 Reff 10V Veff Power consumption = 2units/hour = 2KW = 2000J/s P 2000 I == ; 9A V 220 Power loss in wire = RI2 J/s = ρ lI2 = 1.7 ×10 –8 × 10 × 81 J/sA π×10 –6 � 4 J/s = 0.2% ρ Power loss in Al wire = 4 Al = 1.6 × 4 = 6.4J/s=0.32% ρ Cu 3.30 3.31 4.1 4.2 4.3 4.4 4.5 Let R′ be the resistance of the potentiometer wire. 10 × R′ < 8 ⇒ 10R ′ < 400 + 8R′ 50 + R′ 2R ′ < 400 or R′ < 200Ω. 10 × R′ > 8 ⇒ 2R′> 80 ⇒ R′> 40 10 + R′ 310 × R′ 4 7.5R′ < 80 + 8R′ < 8 ⇒ 10 + R′ R′ > 160 ⇒ 160 < R′ < 200. Any R′ between 160Ω and 200Ω will achieve. Potential drop across 400 cm of wire > 8V. Potential drop across 300 cm of wire < 8V. φ × 400 > 8V (φ→ potential gradient) φ × 300 < 8V φ > 2V/m 2 < 2 V/m.3 6 (a) I = = 1 A = nevd A61 1–4 v = =×10 m/sd 29 –19 –6 10 ×1.6 ×10 ×10 1.6 12KE = 2 me vd × nAl . 11–31 –8 29 –6 –1 –17 × 9.1 10 ××10 ×10 ×10 ×10 ; 2 ×10 J=× 2 2.56 (b) Ohmic loss = RI2 = 6 × 12 = 6 J/s 2×10 –17 –17 All of KE of electrons would be lost in s ; 10 s6 Chapter 4 (d) (a) (a) (d) (a) 124 4.6 4.7 4.8 4.9 4.10 4.11 4.12 4.13 4.14 4.15 4.16 4.17 4.18 4.19 4.20 (d) (a), (b) (b), (d) (b), (c) (b), (c), (d) (a), (b), (d) For a charge particle moving perpendicular to the magnetic field: 2 mv = qvB R qB v ∴ ==ω mR ⎡qB ∴[] ω= ⎢⎣ m dW= F.d l = ⎤⎡ v ⎤ –1 = [] .= T⎥ ⎢⎥⎦⎣ R ⎦ 0 . dt = 0⇒ Fv ⇒ Fv . = 0 F must be velocity dependent which implies that angle between F and v is 90°. If v changes (direction) then (directions) F should also change so that above condition is satisfied. Magnetic force is frame dependent. Net acceleraton arising from this is however frame independent (non - relativistic physics)for inertial frames. Particle will accelerate and decelerate altenatively. So the radius of path in the Dee’s will remain unchanged. At O2, the magnetic field due to I1 is along the y-axis. The second wire is along the y-axis and hence the force is zero. 1ˆˆ0B = ( + +ˆ) µ I ijk 42R ⎡eB ⎤ωNo dimensionless quantity [ ]T –1 = [] = ⎢ ⎥⎣ m ⎦ ˆˆE = E i,E > 0, B = B k0 0 0 Force due to d l2 on d l1 is zero. Force due to d l1 on d l2 is non-zero. 4.21 iG (G + R1) = 2 for 2V range R1R2 R3 iG (G + R1+ R2) = 20 for 20V range and i (G + R+ R+ R) = 200 for 200V rangeG 12 3Gives R1 = 1990Ω R2 = 18 kΩ and R3 = 180 kΩ 4.22 F = BIl Sin θ = BIl 2V 20V 200V µ I B = o 2πh µoI2l F = mg = 2πh QP 2 –7 µoIl 4π×10 × 250 × 25 ×1 h == 2πmg 2π× 2.5 ×10 –3 × 9.8 = 51 × 10–4 h = 0.51 cm 4.23 When the field is off ∑τ= 0 Mgl = Wcoil l 500 g l = Wcoil l Wcoil = 500 × 9.8 N When the magnetic field is switched on Mgl + mgl = Wcoil l + IBL sin 90°l mgl = BIL l 0.2 × 4.9 ×1 10 –2 –3 BIL × m == = 10 kg g 9.8 =1g BV ldB 0F = 1 22 R V ldB 0F = 2 42 R Net τ =τ – τ12 τ= V0 Front view 4.25 As B is along the x axis, for a circular orbit the momenta of the two particles are in the y -z plane. Let p1 and p2 be the momentum of the electron and positron, respectively. Both of them define a circle of radius R. They shall define circles of opposite sense. Let p1 make an angleθ with the y axis p2 must make the same angle. The centres of the repective circles must be perpendicular to the momenta and at a distance R. Let the center of the electron be at Ce and of the positron at Cp. The coordinates of Ce is Side view Z 1.5 R R C e C p B x 4.26 4.27 The coordinates of Ce is Ce ≡ (0,–R sin θ, R cos θ ) The coordinates of Cp is 3 Cp ≡ (0,–R sin θ, R–R cos θ )2 The circles of the two shall not overlap if the distance between the two centers are greater than 2R. Let d be the distance between Cp and Ce. 2 22 ⎛ 3 ⎞Then d = (2RSin ) + ⎜ R – 2Rcos θ ⎟θ ⎝ 2 ⎠ y 2 22 2229 = 4R Sin θ+ R –6R cos θ+ 4R cos θ 4 92 22 = 4R + R –6 R cos θ 4 Since d has to be greater than 2R d2 > 4R2 2922 2⇒ 4R + R –6R cos θ> 4R ⇒ > 6cos θ 4 94 3Or, cos θ< . 8 n = 4 n = 3 n = 2 3 33 Area: A= a 2 A = a2 A = a 2 44 CurrentI is same for all Magnetic moment m = nI A 2I∴m=Ia 23 3a33a 2I (Note: m is in a geometric series) (a)B (z) points in the same direction on z - axis and hence J (L) is a monotonically increasing function of L. (b)J(L) + Contribution from large distance on contour C =µ0I ∴ asL →∞ Contribution from large distance → 0(asB � 1/r 3) J()0∞ − µ I 4.28 4.29 5.1 5.2 5.3 5.4 5.5 5.6 5.7 5.8 5.9 5.10 5.11 µ IR 2 (c) B = 0 z 2 2 3/2 2( z + R ) ∞ ∞µ IR 2 0B dz = dz ∫ z ∫ 2 2 3/2 –∞ –∞ 2( z + R ) Put z = Rtanθ dz= R sec2 θ d θ ∞µ I π /2 ∴ B dz = 0 cos θdθ =µ I∫ z 2 ∫ 0 –∞ –π /2 –L –L (d) B(z) < B (z)squarecircular coil LL∴ℑ ( ) square <ℑ ( ) circular coil But by using arguments as in (b) () =ℑ () ℑ∞ ∞square circular iG .G = (i – i) (S+ S+ S) for i= 10mA1G1231 i (G + S) = (i –i) (S+ S) for i= 100mAG12G232 and i (G + S+S) = (i – i) (S) for i= 1AG123G33 gives S1 = 1W, S2 = 0.1W and S3 = 0.01W (a) zero µ i (b) 0 perpendicular to AO towards left. 2πR (c) µ0 i perpendicular to AO towards left.πR Chapter 5 (c) (a) (c) (b) (b) (a), (d) (a), (d) (a), (d) (a), (c), (d) (b), (c), (d) eheh h µp ≈ µ≈ ,h =and e2mp 2m 2π e µe >> µp because m p >> m e. 5.12 5.13 5.14 5.15 NS 5.16 N S S 5.17 NN 5.18 S 5.19 5.20 Bl =µ Ml =µ (I +I ) and H = 0 = I0 0M Ml = IM = 106 × 0.1 = 105 A. ρN 28g/22.4Lt 3.5 –3 x α density ρ . Now = =×10 = 1.6 × 10–4. ρCu 8g/c c 22.4 x N = 5 ×10 –4 (from given data). xCu Hence major difference is accounted for by density. Diamagnetism is due to orbital motion of electrons developing magnetic moments opposite to applied field and hence is not much affected by temperature. Paramagnetism and ferromagnetism is due to alignments of atomic magnetic moments in the direction of the applied field. As temperature increases, this aligment is disturbed and hence susceptibilities of both decrease as temperature increases. (i)Away from the magnet. z (ii) Magnetic moment is from left to right µ03mr .ˆ ˆB = ,m = mk 4π r 3 yˆ. 2sin θdθds = rr d 0 ≤θ ≤π ,0 ≤φ ≤π ∫ µ0m 3cos θ 2ÑB.ds = ∫ 3 r sin θdθ 4π r x = 0[due to θ integral] . Net m = 0 . Only possibility is shown in Fig. m E (r) = c B (r), p = . Mass and moment of inertia of dipoles are c equal. I 11 m 1T = 2π I ′= ×I and m ′= . T ′= TmB 24 2 2 Consider a line of B through the bar magnet. It must be closed. Let C be the amperian loop. P ∫ P ∫ B H.dl= .dl > 0 µQQ 0 Ñ∫ H.dl=0 PQP 5.21 . 0 Q p d <∫ H l P → Q is inside the bar. Hence H is making an obtuse angle with dl. (i) Along z axis 0 3 2 4 r µ π = m B P S C Q N 0 0 3 2 2 1 11 . 2 –– 4 2 2 R R a a dz m d m z R a µ µ π π ⎛ ⎞⎛ ⎞ = = ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠∫ ∫B l (ii) Along the quarter circle of radius R ( )0 0 0 3 3 ˆ– . – –sin4 4 m B R R µ µ θ π π/ = = m θ z .d =B l 0 2 sin 4 m d R µ θ π 2 0 2 0 B. 4 m dl R π µ π =∫ ur uur (iii) Along x-axis R a1 0 3 – 4 m x µ π ⎛ ⎞ = ⎜ ⎟⎝ ⎠ B O a R x x . 0d =∫ B l (iv) Along the quarter circle of radius a 0 2 – . sin 4 m d d a µ θ θ π =B l , 2 0 0 2 2 0 – – . – 4 4 m m d sin d a a π µ µθ θ π π = =∫ ∫B l Add . 0 C d =∫ B lÑ 5.22 χ is dimensionless. χ depends on magnetic moment induced when H is turned on. H couples to atomic electrons through its charge e. The effect on m is via current I which involves another factor of ‘e’. The combination 2 0 " "µ e does not depend on the “charge” Q dimension. 2 0e m v Rα β γχ µ= 2 2 2 0 2 2 2 0 0 1 1 Energy length ~ . ~ e e c R c c R c µ ε ε = 3 –2 βML T α ⎛ L ⎞ γ 0[ χ ] = M0L0T0Q0 = 2 –2 M ⎜⎟ LQ LT ⎝ T ⎠ α= –1,β= 0,γ= –1 2 –6 –38 µ0 e 10 ×10 χ= ~ ~10–4 .–30 –10 mR 10 ×10 0 m 2 2 1/2 =µ 3 (4cos θ+ sin θ )5.23 (i) B 4π R 2 B 2 = 3cos2 θ+1 θπ , minimum at = .⎛ µ0 ⎞ 2 2m⎜ 3 ⎟⎝ 4π R ⎠ is minimum at magnetic equator.B(ii) tan (dip angle) = BV = 2cot BH π at θ= dip angle vanishes. Magnetic equator is again the2 locus. BV = 1(iii) Dip angle is ± 45° when BH 2 cot θ = 1 θ = tan–12 is the locus. 5.24 Refer to the adjacent Fig. 1.P is in S (needle will point both north) Declination = 0 P is also on magnetic equator. ∴ dip = 0 2.Q is on magnetic equator. ∴ dip = 0 but declination = 11.3°. L L 5.25 n = n = 12π R 24a m = n IA1 m =n IA11 222 L LL2= Iπ R = Ia = Ia 2π R 4a 4 2MR I1 = (moment of inertia about an axis through the diameter)2 2Ma 2I = 12 2 1ω = 1 1 m B I 2 2ω = 2 2 m B I 1m = 2m 1I 2I L 2 LR π × 2 I MR = 2 4 Ia Ma ⇒ a = 3 4 π .R . 2 12 Chapter 6 6.1 6.2 6.3 6.4 6.5 6.6 6.7 6.8 6.9 6.10 6.11 6.12 (c) (b) (a) (d) (a) (b) (a), (b), (d) (a), (b), (c) (a), (d) (b), (c) No part of the wire is moving and so motional e.m.f. is zero. The magnet is stationary and hence the magnetic field does not change with time. This means no electromotive force is produced and hence no current will flow in the circuit. The current will increase. As the wires are pulled apart the flux will leak through the gaps. Lenz’s law demands that induced e.m.f. resist this decrease, which can be done by an increase in current. 132 6.13 6.14 6.15 6.16 y 6.17ˆk x ( ,0,0)a 6.18 BB D P BR v � 6.19 B BQ The current will decrease. As the iron core is inserted in the solenoid, the magnetic field increases and the flux increases. Lent’s law implies that induced e.m.f. should resist this increase, which can be achieved by a decrease in current. No flux was passing through the metal ring initially. When the current is switched on, flux passes through the ring. According to Lenz’s law this increase will be resisted and this can happen if the ring moves away from the solenoid. One can analyse this in more detail (Fig 6.5). If the current in the solenoid is as shown, the flux (downward) increases and this will cause a counterclockwise current (as seen form the top in the ring). As the flow of current is in the opposite direction to that in the solenoid, they will repel each other and the ring will move upward. When the current in the solenoid decreases a current flows in the same direction in the metal ring as in the solenoid. Thus there will be a downward force. This means the ring will remain on the cardboard. The upward reaction of the cardboard on the ring will increase. For the magnet, eddy currents are produced in the metallic pipe. These currents will oppose the motion of the magnet. Therefore magnet’s downward acceleration will be less than the acceleration due to gravity g. On the other hand, an unmagnetised iron bar will not produce eddy currents and will fall with an acceleration g. Thus the magnet will take more time. Flux through the ring φ= Bo (πa 2)cos ωt ε= B(πa 2)ωsin ωt I = B(πa 2)ωsin ωt /R Current at π B (πa 2 )ω t = ; I = along ˆj2ω R π t = ; I = 0 ω 3 π B (πa 2 )ω t = ; I = along –ˆj . 2 ωR One gets the same answer for flux. Flux can be throught of as the number of magnetic field lines passing through the surface (we draw dN = B Δ A lines in an area Δ A ⊥to B), As lines of of B cannot end or start in space (they form closed loops) number of lines passing through surface S1 must be the same as the number of lines passing through the surface S2 . Motional electric field E along the dotted line CD (⊥ to both v and B and along v × B) = vB 6.20 6.21 6.22 6.23 E.M.F. along PQ = (length PQ)×(Field along PQ) Therefore, = cos cos d vB dvB × θ = θ . I dvB R = and is independent of q. Maximum rate of change of current is in AB. So maximum back emf will be obtained between 5s < t <10s. If u = L 1/5 for 3s, dI t dt ⎛ =⎜⎝ 1/5 ⎞ = ⎟⎠ (L is a constant) For 5s < t < 10s 1 3 3 – – –3 5 5 u L L e= = = Thus at t = 7 s, u1 = –3 e. For 10s < t < 30s 2 2 1 20 10 2 L u L e= = = For t > 30s u2 = 0 Mutual inductance 10–2 2 = 5mH = Flux –3 –3 5 10 1 5 10 = × × = × Wb. Let us assume that the parallel wires at are y = 0 and y = d. At t = 0, AB has x=0 and moves with a velocity ˆvi . y At time t, wire is at x (t) = vt. Motional e.m.f. = ( ) ( ˆ)sin – o B t vd ω j E.m.f due to change in field (along OBAC) C A v Total e.m.f – cos ( ) o B t x t d ω ω= ( ) ( )[ ]– cos sin o B d x vt tω ω ω= + O B x Along OBAC, Current (clockwise) ( )cos sin o B d x t v t R ω ω ω= + Force needed along ˆi ( ) sin cos sin o o B d d B tx t v t R ωω ω ω= × ×+ ( ) 2 2 sin .cos sin o B d tx t v t R ωω ω ω= + (i) Let the wire be at x = x (t) at time t. Flux = B (t) l x (t) E – d dt φ = ( ) ( ) ( ) ( ) – – . dB t l x B l v t t t dt = (second term due to motional emf) I 1 E R = Force ( ) ( ) ( ) ( ) ˆ– – l B dB t l x B l v t t t R dt ⎡ ⎤ = ⎢ ⎥⎣ ⎦ i 2 2 d x m dt ( ) 2 2 2 – – l B dB l B dx x t R dt R dt = (ii) dB dt 0, = 2 2 d x dt 2 2l B dx mR dt + 0= C x t( ) D x 2 2 0dv l B v dt mR + = 2 2– A exp l B t v mR ⎛ ⎞ = ⎜ ⎟⎝ ⎠ At t = 0, v = u v (t) = u exp (–l2B2t/mR). (iii) 2 2 2 2 2 2 2 2 2 2 ( ) exp(–2 / )B l v t B l I R R u l B t mR R R = × = Power lost 2 2 2 2 –( )2 2 0 1 – e 2 2 2 t l B t/mR B l mR I R dt u R l B = = ⎡ ⎤⎣ ⎦∫ ( ) 2 2– 2 2 m m u v t= = decrease in kinetic energy. 6.24 Between time t = 0 and 4 t π ω = , the rod OP will make contact with the side BD. Let the length OQ of the contact at some time t such that A 2l B P 0 4 t π ω < < be x. The flux through the area ODQ is 1 1QD×OD= tan 2 2 B B l lφ θ= × CO l D 1 =Bl2 tan θ where θ =ωtP 22lA B dφ 12 2Thus the magnitude of the emf generated is ε= = Bl ω sec ωt dt 2l x l εThe current is I= where R is the resistance of the rod in contact. R C ORD λlR =λx = cos ωt 1B l2ω 2Blω ∴ I = sec ωt cos ωt = 2 λl 2λ cos ωt π 3π For < t < the rod is in contact with the side4ωω A AB. Let the length of the rod in contact (OQ) be x. The 2⎛ 21 l ⎞ P φ= l + Bflux through OQBD is ⎜⎟ where θ = ωt⎝ 2 tan θ ⎠Thus the magnitude of emf generated is dφ 1 sec 2ωt ε= = Bl2ω dt 2 tan 2 ωt εεε sin ωt 1 Bl ω The current is I=== = R λx λl 2 λ sin ωt 3ππ For < t < the rod will be in touch with OC. The Flux throughωω ⎛ l2 ⎞OQABD is φ= 2l 2– B⎜⎟⎝ 2tan ωt ⎠ Thus the magnitude of emf 22dφ Bωl sec ωt ε= = dt 2tans ωt εε 1Blω I== = R λx 2 λ sin ωt 6.25 At a distance r from the wire, dr xµoIField ( ) =Br (out of paper).2πr D r C x o Total flux through the loop is I () tµoI xdr µoI xFlux = l = ln ∫2π r 2π x oxo 6.26 BL1A D x I (t) C L + x2 6.27 1 dI εµ l λ x == I = o ln Rdt R 2π Rx0 If I (t) is the current in the loop. 1 dφ It () = R dt If Q is the charge that passed in time t, dQ dQ 1 dφ I () t = or = dt dt Rdt 1 Integrating Qt ( 1) – Qt ( 2 )= ⎣⎡φ(t1) – φ(t2 )⎦⎤ R L +x µo 2 dx ′ φ(t1)= L1 ∫ I (t1)2π x′ x µ LL + x o 12 = I (t1 ) ln 2π x The magnitute of charge is µ LL + x o 12Q = ln [I –0]2π xo µ LI ⎛ L + x ⎞o 11 2= ln ⎜⎟ . 2π ⎝ x ⎠ 2B.πa2 bE = . . where E is the electric field generated aroundπ EMF = Δt the ring. ⎡ Bπa 2 ⎤Torque = b × Force = QEb = Qb⎢⎥⎣2πbΔt ⎦ 2Ba = Q 2Δt If ΔL is the change in angular momentum 2Ba ΔL = Torque × Δt = Q 2 Initial angular momentum = 0 QBa 2 Final angular momentum = mb 2ω= 2 QBa 2 ω= .2mb 2 d2 xB cos θd ⎛ dx ⎞ 6.28 m 2 = mg sin θ – ⎜⎟ ×(Bd )cos θ dt R ⎝ dt ⎠ 22dv Bd = g sin θ – (cos θ)2 v dt mR 22dv Bd + (cos θ)2 v = g sin θ dt mR g sin θ ⎛ 22 ⎞Bd v =+ A exp ⎜ – (cos 2 θ)t ⎟ (A is a constant to be22 2⎛ Bd cos θ ⎞⎝ mR ⎠⎜⎟⎝ mR ⎠determine by initial conditions) 22mgR sin θ ⎛⎛ Bd ⎞⎞ = 22 2 ⎜1 – exp ⎜ – (cos 2 θ)t ⎟⎟ Bd cos θ ⎝ mR ⎠⎝⎠ 6.29 If Q (t) is charge on the capacitor (note current flows from A to B) vBd Q XAI = – SR RC B v dC B Q dQ vBd ⇒ += YBRC dt R –t /RC Q = vBdC + Ae ∴ ⇒ Q = vBdC [1– e ]–t /RC (At time t = 0, Q = 0 = A = –vBdc). Differentiating, we get vBd –t /RC XAI = e R S BB v d dI L 6.30 –L + vBd = IR BB dt B B B 6.31 6.32 e t dI L + IR = vBd dt vBd – /2 I =+ Ae Rt R vBd At t = 0 I = 0⇒ A = – R I = vBd ( –Rt /L ).1– e R dφ dz = rate of change in flux = (πl2) Bl = IR. dt o dt 2πl B o λ I = v R 2 2 2 2(π l λ) Bo v Energy lost/second = I2 R = R dz This must come from rate of change in PE = mg = mgvdt (as kinetic energy is constant for v= constant) 2 22(π l λB ) vThus, mgv = 0 R mgR Or, v = 2 2. (πl λBo ) Magnetic field due to a solnoid S, B =µ nI 0 Magnetic flux in smaller coil φ= NBA where A =πb2 –dφ –dSo e == (NBA dt dt 2 d(B) 2 d = –N πb = –N πb (µ0dt dt 2 dI = –N πb µ n0 dt 22 2 = –Nn πµ 0bd (mt + C) = –µ0Nn πb 2mt dt e = –µ0Nn πb22mt Negative sign signifies opposite nature of induced emf. The magnitude of emf varies with time as shown in the Fig. 7.1 7.2 7.3 7.4 7.5 7.6 7.7 7.8 7.9 7.10 7.11 7.12 7.13 7.14 7.15 7.16 7.17 7.18 Chapter 7 (b) (c) (c) (b) (c) (c) (a) (a), (d) (c), (d) (a), (b), (d) (a), (b), (c) (c), (d) (a), (d) Magnetic energy analogous to kinetic energy and electrical energy analogous to potential energy. At high frequencies, capacitor ≈ short circuit (low reactance) and inductor ≈ open circuit (high reactance). Therefore, the equivalent circuit Z ≈ R1 + R3 as shown in the Fig. R1 R1 R3 R2 R3 ⇒ (a) Yes, if rms voltage in the two circuits are same then at resonance, the rms current in LCR will be same as that in R circuit. (d) No, because R ≤ a,so Z I ≥ b.I Yes, No. Bandwidth corresponds to frequencies at which m I = 1 2 max I 0.7 max I≈ . 140 7.20 From negative to zero to positive; zero at resonant frequency. 7.21 (a) A (b)Zero (c) L or C or LC 7.22 An a.c current changes direction with the source frequency and the attractive force would average to zero. Thus, the a.c ampere must be defined in terms of some property that is independent of the direction of current. Joule’s heating effect is such property and hence it is used to define rms value of a.c. 7.23 XL = ωL = 2pfL = 3.14Ω ; 3.3Ω ωL tan φ= = 3.14 R φ= tan–1(3.14) ; 72° ; 72 180 × π rad. Timelag tΔ = φ ω 72 180 2 50 π π × = × × = 1 s 250 7.24 7.25 7.26 7.27 PL = 60W, IL = 0.54A 60 VL == 110V. 0.54 1 The transformer is step-down and have input voltage. Hence21i = × I = 0.27A. p2 2 A capacitor does not allow flow of direct current through it as the resistance across the gap is infinite. When an alternating voltage is applied across the capacitor plates, the plates are alternately charged and discharged. The current through the capacitor is a result of this changing voltage (or charge). Thus, a capacitor will pass more current through it if the voltage is changing at a faster rate, i.e. if the frequency of supply is higher. This implies that the reactance offered by a capacitor is less with increasing frequency; it is given by 1/ωC. An inductor opposes flow of current through it by developing a back emf according to Lenz’s law. The induced voltage has a polarity so as to maintain the current at its present value. If the current is decreasing, the polarity of the induced emf will be so as to increase the current and vice versa. Since the induced emf is proportional to the rate of change of current, it will provide greater reactance to the flow of current if the rate of change is faster, i.e. if the frequency is higher. The reactance of an inductor, therefore, is proportional to the frequency, being given by ωL. V2 50,000 Power P = ⇒ = 25 = Z Z 2000 Z2 = R2 + (X – X )2 = 625 C L tan φ = CX L – X 3 –= R 4 625 = 2R 23 – 4 R⎛ ⎞ + ⎜ ⎟⎝ ⎠ = 25 16 R2 = 400 ⇒ R = 20Ω X C – X L = –15Ω I = V = 223 � 9 A. Z 25 MI = 2 9=12.6 A. × 142 If R, XC, XL are all doubled, tan φ does not change. Z is doubled, current is halfed. Power drawn is halfed. 7.28 (i) Resistance of Cu wires, R l 1.7×10 –8 ×20000 =ρ= = 4Ω A ⎛ 1 ⎞2 –4 π × ×10 ⎜⎟⎝ 2 ⎠ 10 6 4I at 220 V: VI = 106 W ; I == 0.45 ×10 A 220 RI2 = Power loss = 4 × (0.45)2 × 108 W > 106 W This method cannot be used for transmission (ii) V′I′ = 106 W = 11000 I′ 1 I′= ×10 2 1.1 21 44RI′= × 4 ×10 =3.3×10 W 1.21 3.3 ×10 4 Fraction of power loss = =3.3% 10 6 v sin ωt m7.29 Ri = v sin ωt i = 1 m 1 R q dq 2 2 + L 22 = v sin ωt Cdt m Let q2 = qm sin (ωt + φ) ⎛ q ⎞ q m – Lω2 sin( ωt +φ) = v sin ωt m ⎜⎟ m⎝ C ⎠ qm = vm ,φ= 0; 1 – ω2L > 012 C– Lω C Rv = 2 – m v Lw 1 2 , Lφ = π ω – 1 C > 0 C 2i = 2dq dt mq= ω cos( )t φω + i1 and i2 are out of phase. Let us assume 1 C 2– Lω 0> 1 2i i+ = m v sin tω R + m v 1 cos tω Lω – cω Now A sin ωt + B cos ωt = C sin (ωt + φ ) C cos φ = A, C sin φ = B; C = A2 + B2 1 ⎡ 22 ⎤2vvmmTherefore, i + i =+ sin( ωt +φ )12 ⎢⎥ ⎣ R2[ωl –1/ ωC]2 ⎦ φ= tan–1 R XL – XC 1/2 1 ⎧ 11 ⎫ += ⎨⎬ Z ⎩R2(Lω –1/ ωC)2 ⎭ di 2 qi di d ⎛ 12 ⎞7.30 Li + Ri += vi ; Li = Li = rate of change of energy stored⎜⎟ dt c dt dt ⎝ 2 ⎠ in an inductor. Ri2 = joule heating loss qd ⎛ q 2 ⎞ i == rate of change of energy stored in the capacitor. ⎜⎟C dt ⎝ 2C ⎠ vi = rate at which driving force pours in energy. It goes into (i) ohmic loss and (ii) increase of stored energy. T TTd ⎛ 12 q 2 ⎞ 2dt + Ri dt = vidt ∫⎜ i + ⎟∫ ∫ 0 dt ⎝ 2 C ⎠ 00 144 7.31 8.1 8.2 8.3 8.4 8.5 8.6 8.7 8.8 8.9 8.10 T 0 + (+ve ) = vidt ∫ 0 T vidt > 0 if phase difference, a constant is acute.∫0 2dq dq q(i) L 2 + R += vm sin ωt dt dt C Let q = qm sin (ω t + φ) = – qm cos (ωt + φ ) i = i msin (ω t + φ ) = qm ω sin (wt + φ ) vm vm –1 ⎛ X – X ⎞i == ;φ= tan ⎜ CL ⎟m 22Z R + (XC – XL ) ⎝ R ⎠ 2 v1 21 ⎡ m ⎤ 2(ii) U = Li = L sin ( ωt +φ)L ⎢ 22 ⎥ 022 ⎢ R + X – X ) ⎥⎣ CL 0 ⎦ 1 q 21 ⎡ vm ⎤21 2UC == ⎢ ⎥ cos ( ωt0 +φ)2 222 C 2CR + (X – X ) ω⎢ CL ⎥⎦⎣ (iii) Left to itself, it is an LC oscillator. The capacitor will go on discharging and all energy will go to L and back and forth. Chapter 8 (c) (b) (b) (d) (d) (c) (c) (a), (d) (a), (b), (c) (b), (d) 8.11 (a), (c), (d) 8.12 (b), (d) 8.13 (a), (c), (d) 8.14 As electromagnetic waves are plane polarised, so the receiving antenna should be parallel to electric/magnetic part of the wave. 8.15 Frequency of the microwave matches the resonant frequency of water molecules. dq 8.16 i = i == –2 πq ν sin2 πν t .CD 0dt 18.17 On decreasing the frequency, reactance X = will increase which c ωC will lead to decrease in conduction current. In this case i D = i C; hence displacement current will decrease. 2 8 –82 1 B 13 ×10 × (12 ×10 )8.18 I = c 0 =× = 1.71 W /m 2. av –6 2 µ0 2 1.26 ×10 E y 8.20 EM waves exert radiation pressure. Tails of comets are due to solar solar radiation. µ 2I µ 1 µ dφ i8.21 0 D 00 EB = ==ε 4πr 4πr 2πr 0 µε d00 2 = (Eπr )2πr dt µε r dE 00= . 2 dt 8.22 (a) λ1 → Microwave, λ3 → X rays, λ4 (b) λ< λ< λ< λ3 2 4 1 dt λ2 → UV → Infrared (c) Microwave - Radar UV - LASIK eye surgery X-ray - Bone fracture identification (bone scanning) Infrared - Optical communication. 8.23 av S 2 0 00c ε ×= E B 2 0 1 cos ( – ) T kx t dt T ω∫ as 2 0( )= c ε ×S E B 2 0 0 0 1 c E B ε= 2 × 2 0 0 0 1 2 E c E c ε ⎛ ⎞ = ×⎜ ⎟⎝ ⎠ 0 0 E asc B ⎛ ⎞ =⎜ ⎟⎝ ⎠ 2 0 0 1 2 ε E c = dV 8.24 iD = C dt dV –3 –6 1×10 = 2 ×10 dt dV 13 =×10 = 5 ×10 V /s dt 2 Hence, applying a varying potential difference of 5 × 102 V/s would produce a displacement current of desired value. 8.25 Pressure Force F 1 Δp ΔpP = == (F == rateofchangeof momentum) Area AA Δt Δt 1 U = .(Δpc =Δ U = energy impartedby wave intime Δt ) A Δtc I ⎛ U ⎞ = ⎜ intensity I = ⎟ c ⎝ AΔt ⎠ 8.26 Intensity is reduced to one fourth. Tis is beacause the light beam spreads, as it propogates into a spherical region of area 4π r2, but LASER does not spread and hence its intensity remains constant. 8.27 Electric field of an EM wave is an oscillating field and so is the electric force caused by it on a charged particle. This electric force averaged over an integral number of cycles is zero since its direction changes every half cycle. Hence, electric field is not responsible for radiation pressure. λ eˆ s ˆ8.28 E = j2πε oa µ i o ˆB= i 2πa µλ v o ˆ= i 2 πa 11 ⎛ λ ˆj µλv ⎞ S= (E× B) = ⎜ s ˆj× o ˆi⎟ µµ 2πε a 2π aoo ⎝ o ⎠ 2–λ v = 4 2 a 2 kˆ πε 0 8.29 Let the distance between the plates be d. Then the electric field oE = V sin(2 πν t ). The conduction current density is given by the Ohm’s d law = E. 1 VoV0⇒ J c = sin (2πν t) = sin(2 πν t)ρ d ρd c= J sin 2πν t o where J0 c = V0. ρd The displacement current density is given as d ∂ E ∂ V J =ε =ε{ o sin(2 πν t)}dt dt d ε 2πν Vo= cos(2 πν t) d d d2πνε V 0 = Jocos(2πν t) , where J0 = d 148 2πνε V ρd = o. = 2πνερ = 2π×80 εν ×0.25 = 4πε ν ×10 d Vo oo 10 ν 4 = = 9 10 9 9× 8.30 (i) Displacement curing density can be found from the relation dEbe JD =ε0 dt ∂ s⎛⎞ ˆ= εµ I cos (2πν t). ln ⎜⎟ k 0 00 ∂ta⎝⎠ 12 ⎛ s ⎞ ˆ= 2 I02πν (– sin (2πν t )) ln ⎜⎟ k c ⎝ a ⎠ 2⎛ν ⎞⎛ a ⎞ ˆ= ⎜⎟ 2π I0 sin (2πν t ) ln⎜⎟k ⎝ c ⎠⎝ s ⎠ 2π ⎛ a ⎞ ˆ= 2 I0 ln ⎜⎟ sin (2πν t ) k λ ⎝ s ⎠ (ii) Id = ∫ JDsdsdθ 2π a ⎛ a ⎞ = 2 I02π ∫ ln ⎜⎟.sds sin (2πν t )λ ⎝ s ⎠ s =0 ⎛ 2π ⎞2 a 12 ⎛ a ⎞ = ⎜⎟ I0 ∫ ds l n⎜⎟.sin (2πν t )⎝ λ ⎠ 2 ⎝ s ⎠ s=0 22 a 22 a ⎛ 2π ⎞⎛ s ⎞⎛ a ⎞ = ⎜⎟ I0 ∫ d ⎜⎟ ln ⎜⎟ .sin (2πν t )4 ⎝ λ ⎠⎝ a ⎠⎝ s ⎠ s=0 2 21 a ⎛ 2π ⎞ = – ⎜⎟ I0 ln ξ d ξ.sin (2πν t )4 ⎝ λ ⎠∫ 0 ⎛ a ⎞2 ⎛ 2π ⎞2 =+ ⎜⎟⎜ ⎟ I0 sin2 πν t ( ∴ The integral has value –1)⎝ 2 ⎠⎝ λ ⎠ (iii) The displacement current 2 d ⎛ a 2π ⎞ dI = ⎜ . I sin2 πν t=I sin 2πν t⎟ 00⎝ 2 λ ⎠ 0 d ⎛ a 2I π ⎞ = ⎜⎟ .I0 ⎝ λ ⎠ x 2341 E.dl = E.dl + E.dl + E.dl + E.dl 8.31 (i) Ñ∫ ∫∫∫∫ E=E î 4 E 3 x1234 dlh dl2 34 1 = E.dl cos 90° + E.dl cos 0 + E.dl cos 90° + E.dl cos 180° ∫ ∫∫ ∫ 1 21 23 4 Zz1 z2dl = E [( – t )–( – ωt )] h sin kz ω sin kz 02 1 B=B0 ˆj (ii) For evaluating ∫ B dsy . let us consider the rectangle 1234 to be made of strips of area ds = h dz each. x Z2 B ds (). = Bds cos0 = Bds = B sin kz – ωt hdz ∫∫ ∫∫ 0 Z1 43 –Bh = [cos( kz – ωt) – cos( kz – ωt)] o 2 1 (2)k dl 2–dφB 1 (iii) ∫ E.dl= z1 z2Ñ dz z B=B ˆj dt yUsing the relations obtained in Equations (1) and (2) and simplifiying, we get y [( – ωt )– ( – ωt)] = Bh ω[( – ωt)– ( – ωt )] Ehsinkz sinkz o sinkz sinkz 021 21k ω E = B00 k E0 = c B0 (iv) For evaluating Ñ∫ B.dl, let us consider the loop 1234 in yz plane as shown in Fig. 2341 B.dl = B.dl + B.dl + B.dl + B.dl Ñ∫ ∫∫∫∫ 1234 234 1 = B dl cos0 + B dl cos 90° + B dl cos 180° + B dl cos90 °∫∫∫ ∫ 123 4 = B h sin kz [( – t )–( – t )] ω sin kz ω (3)01 2 z Now to evaluate φ= ∫E.ds , let us consider the rectangle 1234 toEbe made of strips of area hdz each. Z 2 φ E ds ( ω )= . = Eds cos0 = Eds = E sin kz – t hdz E ∫∫ ∫∫ 01 Z1 –Eh = 0[cos( kz 2– ωt ) – cos( kz 1– ω k dφ Eh ωE0 1 (2 ωt)] ∴= [sin kz (– ωt)– sin kz – dt k (4) ⎛ dφE ⎞In Ñ∫ B dl , I = conduction current. =µ ⎜ I +ε0 ⎟0 ⎝ dt ⎠ = 0 in vacuum. Ñ . =µε 0 dφE∴ ∫ B dl0 dt z Using relations obtained in Equations (3) and (4) and ssimplifying, we get y ω B = E .µε 00 00k E ω 10 = But E0/B0 = c, and ω = ck B0 k µε 00 11 or cc .= Therefore, c = µε 00 128.32 (a) E - field contribution is u =ε EE 20 1 B2 B - field contribution is uB = 2 µ0 1 21 B2 Total energy density u = u + u =ε E + ( 1)EB 022 µ0 The values of E2 and B2 vary from point to point and from moment to moment. Hence, the effective values of E2 and B2 are their time averages. 2 22(E ) =E [sin (kz – ωt )] av 0 av (B 2)= (B 2) = B02[sin 2(kz – ωt )]av av avThe graph of sin2θ and cos2θ are identical in shape but shifted by π/ 2, so the average values of sin2θ and Cos2θ are also equal over any integral multiple of π. and also sin2θ + cos2θ =1 1 So by symmetry the average of sin2θ = average of cos2θ = 2 11222 2∴(E ) = E and B )( = B av 0 av 022 Substuting in Equation (1), 11 B2 u =ε0E2 + 0(2)44 µ 222 2E 11 BE /cE 10 000 2 = c and c = ∴ = =µε =ε E(b) We know 0 00. B0 µε 04 µ04µ04µ00 4 0 111 1222 2Therefore, u =ε E +ε E =ε E ,and I = uc =ε E .av 0000 00 av av 00442 2 Chapter 9 9.1 (a) 9.2 (d) 9.3 (c) 9.4 (b) +1 0 2 9.5 (c) 9.6 (c) 9.7 (b) 9.8 (b) 9.9 (b) 9.10 (d) 9.11 (a) 9.12 (a), (b), (c) 9.13 (d) 9.14 (a), (d) 9.15 (a), (b) 9.16 (a), (b), (c) 9.17 As the refractive index for red is less than that for blue, parallel beams of light incident on a lens will be bent more towards the axis for blue light compared to red. Thus the focal length for blue light will be smaller than that for red. 9.18 The near vision of an average person is 25cm. To view an object with magnification 10, D 25 D m = ⇒ f == = 2.5 = 0.025m fm 10 1 P ==40 diopters. 0.025 9.19 No. The reversibility of the lens makes equation. 9.20 Let the apparent depth be O1 for the object seen from µ2 then µ hO1 = 2 µ13 If seen from µ3 the apparent depth is O2. h µ h µµµ3 ⎛ h ⎞ µ3 ⎛ 2 ⎞ h ⎛ 33 ⎞O =+ O =+=+2 ⎜ 1 ⎟⎜ ⎟⎜⎟3 µ 3 µµµ2 ⎝ 3 ⎠ µ2 ⎝ 1 ⎠ 3 ⎝ 21 ⎠ Seen from outside, the apparent height is 1 ⎛ h ⎞ 1 ⎡hh ⎛ µ3 µ3 ⎞⎤ O3 = ⎜ +O 2 ⎟ = ⎢ + ⎜ + ⎟⎥ µ3 ⎝ 3 ⎠µ3 ⎣ 33 ⎝ µ2 µ1 ⎠⎦ h ⎛ 1 11 ⎞ = ++⎜⎟3 µµ µ⎝ 123 ⎠ 9.21 At minimum deviation ⎡(A + Dm )⎤sin ⎢⎥⎣ 2 ⎦ µ= O ⎛ A ⎞sin ⎜⎟⎝ 2 ⎠ ∴Given D = A m AA2sin cos sin A A22∴µ== = 2cos AA 2sin sin 22 A 3 A ∴ cos = or = 30° ∴ A = 60° 22 2 9.22 Let the two ends of the object be at distance u1= u – L/2 and u2 = u + L/2, respectively, so that |u1–u2|= L. Let the image of the two ends be formed at v1 and v2, so that the image length would be 11 1 fu =. Since + or v = the image of the two ends willL′= v – v12 uv f u – f f (u – L /2 ) f (u + L /2 )v =be at 1 , v2 = u – f – L /2 u – f + L /2 Hence L′=|v – v |= f 2L 212 (u – f )2 × L /4 Since the object is short and kept away from focus, we have L2/4 << (u –f )2 Hence finally 2f L′= L. (u – f )2 9.23 Refering to the Fig., AM is the direction of incidence ray before liquid is filled. After liquid is filledm, BM is the direction of the incident ray. Refracted ray in both cases is same as that along AM. a + Rand sin α= cos(90 – α ) = d2 + (a – R)2 µ(a 2– R2)Substuting, we getd = (a + R)2 – µ(a – R)2 If there was no cut then the object would have been at a height of 0.5 cm from the principal axis 00′. Consider the image for this case. 11 1 – = vu f 1111 11 ∴=+= += vu f –50 25 50 ∴ v = 50 cm. v 50 Magnification is m = =– = –1. u 50 Thus the image would have been formed at 50 cm from the pole and 0.5 cm below the principal axis. Hence with respect to the X axis passing through the edge of the cut lens, the co-ordinates of the image are (50 cm, –1 cm) 9.25 From the reversibility of u and v, as seen from the formula for lens, 1 11 = – f vu It is clear that there are two positions for which there shall be an image on the screen. Let the first position be when the lens is at O. Given –u + v = D ⇒ u = –(D – v) Placing it in the lens formula 1 11 += D –vv f v + D –v 1 ⇒ = (D –vv f) ⇒ v2 – Dv + Df = 0 D D2 –4 Df ⇒ v =± 22 ⎞⎛ DD2–4 Df – ⎟ ⎝ 22 ⎠ Thus, if the object distance is u = –(D – v) = ⎜ ± then the image is at22 22 DD2–4 Df If the object distance is + , then the image is at22 . 22 The distance between the poles for these two object distances is DD2–4 Df ⎛ 2 ⎞DD – 4D f 2+ – ⎟ = D –4 Df ⎜ – 22 ⎝ 22 ⎠ Let d = D2–4 Df Dd Dd If u = + then the image is at v = –.22 22 D–d ∴ The magnification m1 = D + d D– dD+ d If u = then v = 22 D+ dm ⎛ D+ d ⎞2 ∴The magnification m = - Thus 2 = ⎜⎟ . 2D– dm1 ⎝ D– d ⎠ 9.26 Let d be the diameter of the disc. The spot shall be invisible if the dincident rays from the dot at O to the surface at are at the critical angle.2 Let i be the angle of incidence. 1 Then sin i = µ d /2 Now, = tan i hd ⇒ = h tan i = h ⎡⎣2 2h ∴ d = 9.27 (i) Let the power at the far point be Pf for the normal relaxed eye. 11 1Then P = =+ = 60 D ff 0.1 0.02 With the corrective lens the object distance at the far point is ∞. The power required is 11 1 P′==+ = 50D ff ′∞ 0.02 The effective power of the relaxed eye with glasses is the sum of the eye and that of the glasses Pg . ∴ P ′ = P + Pf fg ∴ P = – 10 D. g(ii) His power of accomadation is 4 diopters for the normal eye. Let the power of the normal eye for near vision be Pn . Then 4 = P – P or P = 64 D. nf n Let his near point be x n, then 11 1 += 64 or + 50 = 64 x 0.02 x nn 1 = 14, x n 1 ∴ x = ; 0.07m n14 (iii) With glasses Pn ′= Pf ′+ 4 = 54 11 154 =+ =+ 50 xn ′ 0.02 xn ′ 1 = 4, x′ n 1 ∴ xn ′= = 0.25m .4 9.28 Any ray entering at an angle i shall be guided along AC if the angle the ray makes with the face AC ( φ ) is greater than the critical angle. 1A C ⇒ sin ≥ µ 1 r ⇒ cos r ≥ i µ 1Or, 1 – cos2r ≤ 1 – 2 BDµ 1i.e. sin2r ≤ 1 – 2 µ Since sin i = µ sin r 1 1 2 sin2i ≤ 1 – 2µ µ Or, sin2i ≤µ2 – 1 π The smallest angle φ shall be when i = . If that is greater than the 2 critical angle then all other angles of incidence shall be more than the critical angle. Thus 1 ≤µ2 –1 Or, µ2 ≥ 2 ⇒µ≥ 2 9.29 Consider a portion of a ray between x and x + dx inside the liquid. Let the angle of incidence at x be θ and let it enter the thin column at height y. Because of the bending it shall emerge at x + dx with an angle θ + dθ and at a height y + dy. From Snell’s law µ(y) sin θ = µ(y+dy) sin (θ+dθ) ⎛ dµ ⎞ µ yor µ(y) sinθ ; ⎜ () + dy ⎟ (sinθ cosdθ + cosθ sin dθ )dy ⎝⎠ d µ y sin θ; µ( )sin θ+µ( )cos y θdθ+ dy dy dy � + d� –dµ (y + dy)or µ(y) cosθdθ ; dy sin θ dy –1 dµdθ ; dy tan θ µ dy dx But tanθ = (from the fig.)dy r + dr –1 dµ∴ dθ = dx µ dy –1 dµ d –1 dµ∴θ = dx = d∫µ dy o µ dy 9.30 Consider two planes at r and r + dr. Let the light be incident at an angle θ at the plane at r and leave r + dr at an angle θ +dθ Then from Snell’s law n(r) sinθ = n(r + dr) sin (θ + dθ ) ⎛ dn ⎞ ⇒ n(r) sinθ ; ⎜nr () + dr ⎟ (sinθ cos dθ + cosθ sin dθ )⎝ dr ⎠ ⎛ dn ⎞;⎜nr () + dr ⎟ (sinθ + cosθ dθ )⎝ dr ⎠ Neglecting products of differentials () θ ; () θ+ dn dr sinθ + n(r) cosθdθnrsin nrsin dr dn dθ ⇒ – tan θ= nr () dr dr 2GM ⎛ 2GM ⎞ dθ dθ tan θ= 1+≈⇒ 22 ⎜⎟ rc ⎝ rc 2 ⎠ dr dr r θo ∞2GM tan θdr ∴ dθ=∫ 2 ∫ 2 cr0–∞ RNow r2 = x2 +R2 and tanθ = x 2rdr = 2xdx θo ∞2GM R xdx dθ=∫ 2 ∫ 3 cx0–∞ 22(x + R )2 Put x = R tan φ dx = R Sec2 φ d φ π /2 22GMR R sec φ dφ ∴θ = 0 2 ∫ 33c R sec φ –π /2 π /2 2GM 4GM = cos φ d φ = 2 ∫ 2Rc Rc –π /2 9.31 As the material is of refractive index –1, θ r is negative and θ′ positive.r Now = =θ r ′θi θ r The total deviation of the outcoming ray from the incoming ray is 4θ i . Rays shall not reach the receiving plate if π 3π ≤ 4θi ≤ (angles measured clockwise from the y axis)22 π 3π ≤θi ≤ y Receiving plate88 x Now sin θi = R �i R π x 3π ≤ sin –1 ≤� xr r8 R 8 π x 3π Or, ≤≤ 8 R 8 x Rπ R3πThus for ≤ x ≤ light emitted from the source shall not reach88 the receiving plate. 9.32 (i) The time required to travel from S to P1 is 2SP u 2 + b2 u ⎛ 1 b ⎞ t1 = 1 = ;⎜1+ 2 ⎟ assuming b << u0c cc ⎝ 2 u ⎠ The time required to travel from P1 to O is 22 2PO v + bv ⎛ 1 b ⎞ t2 = 1 = ; ⎜1+ ⎟ c cc ⎝ 2 v 2 ⎠ The time required to travel through the lens is (n –1) () t =wb where n is the refractive index. l c Thus the total time is 1 ⎡ 12 ⎛ 11 ⎞⎤ 1 11 P1t = u + v + b ++ (n –1) () wb ⎢⎜⎟ ⎥ . Put =+ c ⎣ 2 ⎝ uv ⎠⎦ Duv 22 S1 ⎛ 1 b ⎛ b ⎞⎞Then t = ⎜u + v ++ (n –1) ⎜w + ⎟⎟ c 2 D 0⎝⎝ α ⎠⎠ Fermet’s principle gives dt b 2( n –1) b = 0 = – db CD cα α= 2( n –1) D Thus a convergent lens is formed if α= 2( n –1) D . This is independant of b and hence all paraxial rays from S will converge at O (i.e. for rays b << n and b << v). 1 11Since , the focal length is D.=+ Duv (ii) In this case 1 ⎛ 1 b2 ⎛ k2 ⎞⎞ u + v + +()k ln t = ⎜ n −11 ⎜ ⎟⎟ c ⎝ 2D ⎝ b ⎠⎠ dt b k = 0 = –(n –1) 1 db D b ⇒ b2 = (n – 1) k1D ∴ b = (n –1) kD 1 Thus all rays passing at a height b shall contribute to the image. The ray paths make an angle b (n –1) kD (n –1) k uv (n –1) ku 11 1β ; = = = . vv 2 v 2(u + v)(u + )vv Chapter 10 10.1 (c) 10.2 (a) 10.3 (a) 10.4 (c) 10.5 (d) 10.6 (a), (b), (d) 10.7 (b), (d) 10.8 (a), (b) 10.9 (a), (b) 10.10 Yes. 10.11 Spherical. 10.12 Spherical with huge radius as compared to the earth’s radius so that it is almost a plane. 10.13 Sound wave have frequencies 20 Hz to 20 kHz. The corresponding wavelengths are 15m and 15mm, respectively. Diffraction effects are seen if there are slits of width a such that. α : λ. For light waves, wavelengths � 10–7m. Thus diffraction effects will show when a : 10–7 m. whereas for sound they will show for 15mm < a < 15m . 2.54 10.14 The linear distance between two dots is l = cm ; 0.84 ×10 –2 cm. 300 At a distance of Z cm this subtends an angle. l 0.84 ×10 –2 cm φ : l /z ∴ z == –4 : 14.5cm . φ 5.8 ×10 10.15 Only in the special cases when the pass axis of (III) is parollel to (I) or (II) there shall be no light emerging. In all other cases there shall be light emerging because the pass axis of (II) is no longer perpendicular to the pass axis of (III). 10.16 Polarisation by reflection occurs when the angle of incidence is the nBrewster’s angle i.e. tan θB = 2 where n2 < n1. n1 When light travels in such a medium the critical angle is where n2 < n1 . As |tan θ|>|sin θ |for large angles, θ <θ .Bc BC Thus, polarisation by reflection shall definitely occur. 1.22 λ d =10.17 min 2sin β nsin θ= 2 c n1 where β is the angle subtended by the objective at the object. o For light of 5500 A 1.22 × 5.5 ×10 –7 d = mmin 2sin β For electrons accelerated through 100V the deBroglie wavelength is h 1.227 –9 λ== = 0.13nm = 0.13 ×10 m p 100 1.22 ×1.3 ×10 –10 ∴d ' = min 2sin β 162 1.22 ×1.3 ×10 –10 ∴d 'min = 2sin β d ' 1.3 ×10 –10 min –3 = –7 : 0.2 ×10 dmin 5.5 ×10 10.18 T2P = D + x, T1 P = D – x = [D2 + (D – x)2]1/2 S2P = [D 2 + (D + x)2]1/2 Minima will occur when λ[D 2 + (D + x)2]1/2 – [D2 + (D – x)2]1/2 = 2 If x = D λ (D2 + 4D2)1/2 = 2 λ (5D2)1/2 = λ , ∴ D = .2 25 10.19 Without P: A = A⊥+ A11 120 0A = A + A = A sin( kx –ωt )+A sin( kx – ωt +φ )⊥⊥⊥⊥ ⊥ (1) (2) A = A + A11 11 11 A = A0 [sin( kx – wt ) + sin( kx – ωt +φ]11 11 where A0 ⊥ , 0 are the amplitudes of either of the beam in ⊥ andA1111 polarizations. ∴ Intensity = 2 22 222A0 A0 } [sin (kx – wt )(1+cos φ+ 2sin φ )+sin ( kx –ωt) sin φ] average +={ ⊥ 11 2 2 ⎛ 1 ⎞0 0 }⎜⎟.2(1+cos φ)+={ A⊥ A11 ⎝ 2 ⎠ 02 A0 ⊥ 0.(1 + cos φ )since =A⊥= 2 A11 average average With P: Assume A2 ⊥ is blocked: 1 22 12Intensity = (A + A) + (A )11 11 ⊥ 2 21 A0 A0= (1 + cos φ) + .⊥ ⊥ 2 02Given: I0 = 4 = Intensity without polariser at principal maxima. Intensity at principal maxima with polariser 02 ⎛ 1 ⎞ A⊥ = ⎜2 + ⎟A⊥ ⎝ 2 ⎠ 5 = I08 Intensity at first minima with polariser 02 02 A⊥ = A⊥ (1–1)+ 2 I0= .8 10.20 Path difference = 2d sin θ+ (µ –1) l ∴ For principal maxima, 2d sin θ+ 0.5 l = 0 –l –1 ⎛ d ⎞ 0 ⎜⎟sin θ= = Q l = 4d 16 ⎝ 4 ⎠ DOP = D tan θ≈ –∴ 0 16 For the first minima: λ2d sin θ +0.5 l = ±∴ 1 2 164±λ /2 – 0.5 l ±λ /2 – λ /8 11sin θ1 = = =±– 2d 2λ 4 16 3 θ+On the positive side: sin = 16 –5 On the negative side: sin θ= – 16 The first principal maxima on the positive side is at distance sin θ+ 3 –5 In the –ve side, the distance will be D tan θ= below O. 16 2 –52 10.21 (i) Consider the disturbances at R1 which is a distance d from A. Let the wave at R1 because of A be YA = a cos ωt. The path difference of the signal from A with that from B is λ/2 and hence the phase difference is π. Thus the wave at R1 because of B is y = a cos( t – π ) = –a cos ωt.B ω The path difference of the signal from C with that from A is λ andR2 hence the phase difference is 2π. Thus the wave at R1 because of C is yc = a cos ωt. The path difference between the signal from D with that of A is 2 ⎛ λ ⎞2 d + ⎜⎟ −(d −λ /2 )/2 /2 ⎝ 2 ⎠ R1 A BC 1/2 ⎛ λ ⎞ λ = d 1+− d +⎜⎟/2 2⎝ 4d ⎠ 2 D 2 1/2 ⎛ λ ⎞ λ = d 1+− d +⎜ 2 ⎟⎝ 8d ⎠ 2 If d >>λ the path difference : λ and hence the phase difference2 is π. ∴yD =− a cos ωt . Thus, the signal picked up at R1 is y + y+ y+ y= 0 AB C D Let the signal picked up at R from B be y= a cos ωt. 2B 1The path difference between signal at D and that at B is λ/2. ∴yD = –a cos ωt1 The path difference between signal at A and that at B is 2 1/2 2 ⎛ λ ⎞⎛ λ2 ⎞ 1 λ2 () + ⎜⎟ − d = d 1+− d :d ⎜ 2 ⎟ 2⎝ 2 ⎠⎝ 4d ⎠ 8 d 2πλ2 πλ ∴ The phase difference is . = =φ : 0. 8λ d24d Hence, yA = a1 cos (ωt-φ) Similarly, yC = a1 cos (ωt-φ) ∴ Signal picked up by R2 is y+ y+ y+ y= y = 2acos (ωt-φ)A BCD 1 2 22∴|| = 4a1 cos ( ωt –y φ ) = 2a 2∴ I 1 Thus R1 picks up the larger signal. (ii) If B is switched off, R1 picks up y = a cos ω t 12∴ = aIR1 2 R2 picks up y = a cos ω t 12 = a 21∴ IR2 Thus R1 and R2 pick up the same signal. (c) If D is switched off. R1 picks up y = a cos ω t 12∴ = aIR1 2 R2 picks up y = 3a cos ω t 12∴ = 9aIR2 2 Thus R2 picks up larger signal compared to R1 . (iv)Thus a signal at R1 indicates B has been switched off and an enhanced signal at R2 indicates D has been switched off. 10.22 Fig.1 Fig. 2 10.23 (i) Suppose the postulate is true, then two parallel rays would proceed as shown in Fig. 1. Assuming ED shows a wave front then all points on this must have the same phase. All points with the same optical path length must have the same phase. Thus – εµ AE = BC – εµ CD rr rr or BC = εµ ( )r rCD − AE BC > 0, CD > AE As showing that the postulate is reasonable. If however, the light proceeded in the sense it does for ordinary material (viz. in the fourth quadrant, Fig. 2) Then – εµ AE = BC – εµ CD rr rr or, BC = εµ ( )r rCD − AE As AE > CD, BC < O showing that this is not possible. Hence the postalate is correct. (ii) From Fig. 1. BC = AC sin θi and CD-AE = AC sin θ r: Since − εµ ( CD )= BC r rAE −–n sin θ r = sin θ i . Consider a ray incident at an angle i. A part of this ray is reflected from the air-film interface and a part refracted inside. This is partly reflected at the film-glass interface and a part transmitted. A part of the reflected ray is reflected at the film-air interface and a part transmitted as r2 parallel to r1 . Of course succesive reflections and transmissions will keep on decreasing the amplitude of the wave. Hence rays r1 and r2 shall dominate the behavior. If incident light is to be transmitted through the lens, r1 and r2 should interfere destructively. Both the reflections at A and D are from lower to higher refractive index and hence there is no phase change on reflection. The optical path difference between r2 and r1 is n (AD + CD) – AB. If d is the thickness of the film, then d AD = CD = cos r AB = AC sin i AC = d tan r 2 ∴ AC = 2d tan r Hence, AB = 2d tanr sini Thus the optical path difference is 11.1 11.2 11.3 11.4 11.5 11.6 11.7 11.8 11.9 11.10 11.11 d2n − 2d tan r sin i cos r r1 sin id sin r = 2. − 2d sin i sin r cos r cos r Air Film⎡ 1− sin 2 r ⎤ = 2d sin ⎢⎥ Dsin r cos r Glass n = 1.5⎣⎦ = 2nd cosr For these waves to interefere destructively this must be λ/2. λ⇒ 2nd cos r = 2 or nd cos r = λ/4 For a camera lens, the sources are in the vertical plane and hence i � r � 0 λ ∴nd ; .4 o o5500 A ⇒ d = ;1000 A 1.38 × 4 Chapter 11 (d) (b) (d) (c) (b) (a) (a) (c) (c), (d) (a), (c) (b), (c) 11.12 (a), (b), (c) 11.13 (b), (d) 2mE λ /λ= p /p = αα= 8 :1 11.14 pd xp 2mE pp 11.15 (i) E max = 2hν – φ (ii) The probability of absorbing 2 photons by the same electron is very low. Hence such emissions will be negligible. 11.16 In the first case energy given out is less than the energy supplied. In the second case, the material has to supply the energy as the emitted photon has more energy. This cannot happen for stable substances. 11.17 No, most electrons get scattered into the metal. Only a few come out of the surface of the metal. 11.18 Total E is constant Let n1 and n2 be the number of photons of X-rays and visible region nE = nE 11 2 2 hc hc n = n12λλ12 n λ11= . n λ n 1 22 1 = . n2 500 11.19 The momentum is transferred to the metal. At the microscopic level, atoms absorb the photon and its momentum is transferred mainly to the nucleus and electrons. The excited electron is emitted. Conservation of momentum needs to be accounted for the momentum transferred to the nucleus and electrons. 11.20 Maximum energy = hν – φ ⎛1230 ⎞ 1 ⎛1230 ⎞⎜ – φ ⎟ = ⎜ – φ ⎟⎝ 600 ⎠ 2 ⎝ 400 ⎠ 1230 φ= = 1.02eV. 1200 11.21 ΔxΔp ;h h 1.05 ×10 –34 Js –25 Δp ; ; = 1.05 ×10 Δx 10 –9 m 2 –252 2 2 p (1.05 ×10 ) 1.05 1.05 E == =×10 –19 J= eV 2m 2 × 9.1 ×10 –31 18.2 18.2 ×1.6 = 3.8 × 10–2eV 11.22 I = nn = n ν AA B B nA ν B= 2 = nB ν A The frequency of beam B is twice that of A. hhhh p + p11.23 pc = = + == if p, p > 0 or p, p< 0A B λλ λλ ABAB AB c c λλ λ= AB or c λA +λB If p> 0, p < 0 or p< 0, p> 0 A BA B λB – λAh pc = h = λA .λB λ c λB.λAλ= . c λA – λB 11.24 2d sinθ = λ = d =10–10 m. h 6.6 ×10 –34 –21 p == = 6.6 ×10 kgm/s –10 –10 10 10 –242 2 –19 –2 (6.6 ×10 ) 6.6 E =×1.6 ×10 =×1.6 ×10 eV 2× (1.7 ×10 –27 ) 2×1.7 = 20.5 × 10–2eV = 0.21eV 11.25 6 × 1026 Na atoms weighs 23 kg. Volume of target = (10–4 × 10–3) = 10–7m3 Density of sodium = (d) = 0.97 kg/m3 23 3Volume of 6 × 1026 Na atoms = m = 23.7 m3 0.9723 3Volume occupied of 1 Na atom = 26 m = 3.95 × 10–26 m3 0.97 × 6 ×10 10 –7 No. of sodium atoms in the target = –26 = 2.53 × 1018 3.95 ×10Number of photons/s in the beam for 10–4 m2 = n 170 Energy per s nhν = 10–4 J × 100 = 10–2 W 1234.5 hν (for λ = 660nm) = 600 = 2.05eV = 2.05 × 1.6 × 10–19 = 3.28 × 10–19J. 10 –2 16 n = = 3.05 ×10 /s 3.28 ×10 –19 1 17 16 n = 10 = 3.1 10 ×× 3.2 If P is the probability of emission per atom, per photon, the number of photoelectrons emitted/second 16 18 = P × 3.1 10 × 2.53 ××10 Current = P × 3.1 × 10+16 × 2.53 × 1018 × 1.6 × 10–19 A = P × 1.25 × 10+16 A This must equal 100µA or 100 ×10 –6 P = 1.25 ×10+16 ∴ P = 8 × 10–21 Thus the probability of photemission by a single photon on a single atom is very much less than 1. (That is why absorption of two photons by an atom is negligible). ∞ 2211 q 1 q11.26 Work done by an external agency = + . dx = .∫ 24πε 04 dx 44πε 0d –19 9(1.6 ×10 )× 9 ×10 With d = 0.1nm, energy = –10 –19 eV 4(10 )×1.6 ×10 1.6 × 9 = eV = 3.6 eV4 11.27 (i) Stopping potential = 0 at a higher frequency for B. Hence it has a higher work function. h 2 (ii) Slope = = 14 for A. e(10 – 5)×102.5 = 14 for B.(15 –10)×101.6 ×10 –19 –14 –34 h =× 2×10 = 6.04 ×10 JsforA 5 –19 –14 1.6 ×10 × 2.5 ×10 = = 8 × 10–34 Js for B.5 Since h works out differently, experiment is not consistent with the theory. 11.28 mv = mv + mv A A1 B2 1 2 1 21 2 mv = mv + mvA A 1 B 2222 1/ 1/ 2∴ m (v – v )( v + v ) = mvA 1 A 1 BB2/ 2/ ∴ v + v = v 12 or v = v – v 21 ⎛ m – m ⎞ ⎛ 2m ⎞AB A∴v = v, and v = v1 ⎜⎟ 2 ⎜⎟ m + m m + m⎝ AB ⎠ ⎝ AB ⎠ h ∴λ = initial mvA ( A + mB )hm h λ= = final m (m – m )vmv AA BA mA + mB ) ⎤h ⎡ ∴Δ λ = –1⎢ ⎥ m – mmv ABA ⎣ ⎦ dN P 11.29 (i) == 5 ×10 19 /se dt (hc /λ) (ii) hc = 2.49eV > W : Yes. 0 dλ 2πr(iii) P. Δt = W0, Δt = 28.4s 4πd2 ⎛ dN ⎞ πr 2 (iv) N = ⎜⎟ × 2 ×Δ t = 2 ⎝ dt ⎠ 4π d Chapter 12 12.1 (c) 12.2 (c) 12.3 (a) 12.4 (a) 12.5 (a) 12.6 (a) 12.7 (a) 12.8 (a), (c) 12.9 (a), (b) 12.10 (a), (b) 12.11 (b), (d) 12.12 (b), (d) 12.13 (c), (d) 12.14 Einstein’s mass-energy equivalence gives E = mc2. Thus the mass of a H-atom is m + m −B where B ≈ 13.6eV is the binding energy.pe 2 c 12.15 Because both the nuclei are very heavy as compared to electron mass. 12.16 Because electrons interact only electromagnetically. 12.17 Yes, since the Bohr formula involves only the product of the charges. 13.6 12.18 No, because accoding to Bohr model, E = –,n 2 n and electons having different energies belong to different levels having different values of n. So, their angular momenta will be nh different, as mvr = . 2π 4 12.19 The ‘m’ that occurs in the Bohr formula E =– me is the n 228ε0nh reduced mass. For H-atom m ≈ m e. For positronium m ≈ me /2 . Hence for a positonium E1≈ – 6.8eV. 12.20 For a nucleus with charge 2e and electrons of charge –e, the levels 4me 4 E = – are n 2 2 2 . The ground state will have two electrons each8ε0 nh of energy E, and the total ground state energy would by –(4×13.6)eV. 12.21 v = velocity of electron a0 = Bohr radius. 2πa0∴Number of revolutions per unit time = v 2πa0∴ Current = e. v 2 ⎡ 11 ⎤ 12.22 ν= cRZ 2 − 2,mn ⎢⎥(n + p ) n⎣⎦ where m = n + p, (p = 1, 2, 3, ...) and R is Rydberg constant. For p << n. 2 ⎡ 1 ⎛ p ⎞–2 1 ⎤ ν= cRZ mn ⎢ 2 ⎜1+ ⎟ − 2 ⎥ ⎣n ⎝ n ⎠ n ⎦ 2 ⎡ 12p 1 ⎤ν= cRZ – −⎢ 232 ⎥mn ⎣n nn ⎦ 22p ⎛ 2cRZ 2 ⎞ν= cRZ ; pmn 3 ⎜⎟ n ⎝ n 3 ⎠ Thus, ν mn are approximately in the order 1, 2, 3........... 12.23 Hγ in Balmer series corresponds to transition n = 5 to n = 2. So the electron in ground state n = 1 must first be put in state n = 5. Energy required = E1 – E5 = 13.6 – 0.54 = 13.06 eV. If angular momentum is conserved, angular momentum of photon = change in angular momentum of electron = L5– L2 = 5h –2h = 3h = 3×1.06 ×10 –34 = 3.18 × 10–34 kg m2/s. me ⎛ me ⎞12.24 Reduced mass for H =µ = ; m 1– H e ⎜⎟ e ⎝ M ⎠1+ m M 174 ⎛ me ⎞⎛ me ⎞⎛ me ⎞Reduced mass for D =µD ; me ⎜1– ⎟ = me ⎜1– ⎟⎜ 1 + ⎟⎝ 2M ⎠⎝ 2M ⎠⎝ 2M ⎠ 1 hν = (E – E )αµ. Thus, λα ij ij ij µ If for Hydrogen/Deuterium the wavelength is λH /λD λD µH ⎛ m ⎞–1 ⎛ 1 ⎞ = ; ⎜1+ e ⎟ ; ⎜1– ⎟λH µD ⎝ 2M ⎠⎝ 2 ×1840 ⎠ λD =λH × (0.99973) Thus lines are 1217.7 Å , 1027.7 Å, 974.04 Å, 951.143 Å. 12.25 Taking into account the nuclear motion, the stationary state µZ 2 e 4 ⎛ 1 ⎞energies shall be, En = – ⎜⎟ . Let µH be the reduced mass22 28ε0h ⎝ n ⎠ of Hydrogen and µD that of Deutrium. Then the frequency of the 4 4 1st µHe ⎛ 1 ⎞ 3 µHeLyman line in Hydrogen is hν= 1– = . ThusH 22 ⎜⎟ 228ε0 h ⎝ 4 ⎠ 48ε0h3 µHe 4 the wavelength of the transition is λ= . The wavelengthH 2348ε0hc 3 µDe 4 of the transition for the same line in Deutrium is λ= .D 2348ε0hc ∴Δ λ =λD – λH Hence the percentage difference is Δλ λ – λµ – µD H DH100 ×= ×100 =×100 λλ µHH H m M mM e D eH– (m + M )(m + M )eD eH= ×100 mM /( m + M )e He H ⎡⎛ m + M ⎞ M ⎤e HD= –1 ×100 ⎢⎜ ⎟⎥ m + MM⎣⎝ eD ⎠ H ⎦ Since m << M < M eHD Δλ ⎡ MM ⎛1+ m /M ⎞⎤HD eH×100 =× –1 ×100 ⎢⎜ ⎟⎥λ MM 1+ m /MH ⎣ DH ⎝ eD ⎠⎦ = ⎡⎣(1 + me /MH )(1 + me /MD )–1 –1⎤⎦ ×100 mm⎡ ee ⎤ ; (1 + – –1 ×100 ⎢ ⎥MM⎣ HD ⎦ ⎡ 11 ⎤ ≈ m – ×100 e ⎢⎥MM⎣ HD ⎦ –31 ⎡ 11 ⎤ = 9.1 10 ×100 × – ⎢ –27 –27 ⎥⎣1.6725 ×10 3.3374 ×10 ⎦ =× [ – 0.2996 ]×100 9.1 10–4 0.5979 = 2.714 × 10–2 % 12.26 For a point nucleus in H-atom: mv 2 e 21Ground state:mvr = h, = –. rr 24πε BB 0 h21 ⎛ e 2 ⎞ 1 ∴m . =+ ⎜2 2 ⎟ 2 mr r 4πε rB B ⎝ 0 ⎠ B h2 4πε ∴ . 20 = rB = 0.51A ° me Potential energy 2 22⎛ e ⎞ 1 mv 1 hh –⎜⎟. = –27.2 eV KE ;. == m. = =+ 13.6eV 22 24πrr 22 mr 2mr ⎝ 0 ⎠ B BB For an spherical nucleus of radius R, If R < rB, same result. If R >> rB: the electron moves inside the sphere with radius rB ′ (rB ′ = new Bohr radius). 4 ⎛ r′3 ⎞Charge inside r′= eB B ⎜⎟⎝ R3 ⎠ 176 h2 ⎛ 4πε 0 ⎞ R3 B ⎜ 2 ⎟ 3∴r′= m ⎝ e ⎠ rB ′ ° r′ B 4 = (0.51A). R3. R = 10 A ° ° = 510(A)4 ∴rB ′≈ (510)1/4 A °< R. 12 m hh 1 . = mv = . = .KE 22 222 m r B ′ 2m rB ′ 22 2⎛ h ⎞⎛ r ⎞ (0.51) 3.54 = . B = (13.6eV) == 0.16eV ⎜ 2 ⎟⎜ 2 ⎟ 1/2 2mr r′ (510) 22.6 ⎝ B ⎠⎝ B ⎠ 22⎛ e 2 ⎞⎛ r′ –3R ⎞ . =+ ⎜ . BPE ⎟⎜ 3 ⎟ ⎝ 4πε 0 ⎠⎝ 2R ⎠ ⎣ 1000 ⎦ –141 =+ (27.2eV). = –3.83eV. 1000 12.27 As the nucleus is massive, recoil momentum of the atom may be neglected and the entire energy of the transition may be considered transferred to the Auger electron. As there is a single valence electron in Cr, the energy states may be thought of as given by the Bohr model. The energy of the nth state En =–Z 2R 12 where R is the Rydberg n constant and Z = 24. The energy released in a transition from 2 to 1 2 ⎛ 1 ⎞ 32is ΔE = ZR ⎜1– ⎟ = ZR. The energy required to eject a n = 4 ⎝ 4 ⎠ 4 21electron is E4 = ZR . 16 Thus the kinetic energy of the Auger electron is 2 ⎛ 31 ⎞ 12 . = ZR –KE ⎜⎟ = ZR ⎝ 4 16 ⎠ 16 11 =× 24 × 24 ×13.6eV 16 = 5385.6 eV 12.28 mc2 = 10–6 × electron mass × c2 p ≈ 10–6 × 0.5MeV –6 –13 ≈ 10 × 0.5 ×1.6 ×10 ≈ 0.8 ×10 –19 J –34 8hhc 10 × 3 ×10 2 –19 ≈ 4 ×10 –7 m >> Bohr radius. == mc mc 0.8 ×10 pp e 2 ⎡ 1 λ ⎤ = 2 + exp(– λr )F ⎢⎥4πε 0 ⎣rr ⎦ –1 –7 where λ= h ≈ 4 ×10 m >> rB mc p 1 << . λ B << 1∴λ ie r rB e 2 exp(– λr)Ur () = –. 4πε 0 r h mvr = h∴ v = mr mv 2 ⎛ e 2 ⎞⎡ 1 λ ⎤Also : =≈ +⎜ ⎟⎢ 2 ⎥r ⎝ 4πε 0 ⎠⎣rr ⎦ h2 ⎛ e 2 ⎞⎡ 1 λ ⎤∴ 3 = ⎜ ⎟⎢ 2 + ⎥mr 4πε ⎣rr ⎦⎝ 0 ⎠ h 2 ⎛ e 2 ⎞ 2∴= ⎜⎟ [r +λr ] m 4πε ⎝ 0 ⎠ 178 h 4πε If λ= 0; r = rB = . 20 me 22h e = .r m 4πε 0 B Since λ –1 >> rB ,put r = rB +δ 22 2∴ r = r +δ +λ(r +δ + 2δr );negect δBBB B or 0 =λrB 2 +δ (1 + 2λrB ) 2 B 22 –λ r δ= ≈λ r (1– 2λ r ) = –λr since λr << 1BBB B1+ 2λ rB e 2 exp(– λδ – λrB )∴Vr () = –. 4πε 0 rB +δ e 21 ⎡ δ ⎤ ∴Vr () = – ⎢⎜⎛1– ⎟⎞.(1 – λrB )⎥4πε r ⎝ B ⎠0 B ⎣ r ⎦ ≅ (–27.2eV)remains unchanged. 2221 21 hhh ⎛ 2δ ⎞KE = mv –.– = m. == 12 22 ⎜⎟22 mr 2( rB +δ )2rB ⎝ rB ⎠ = (13.6eV) 1 + 2λr e 2 h2 Total energy = – + 2 [1+ 2λrB ] [ B ] 4πε r 2r0 BB + [+ 2λr ] eV = –27.2 13.6 1 B Change in energy = 13.6 × 2λrBeV = 27.2 λrBeV 12.29 Let ε= 2 +δ qq Rδ Rδ 120 0 qq –19 2 F = .2+δ =∧ 2+δ , where 1 2 =∧ , ∧= (1.6 ×10 ) × 9×10 9 4πε rr 4πε 00 = 23.04 ×10 –29 2 mv = r ∧Rδ 20 v = 1+δ mr 1/2 nh nh ⎡ m ⎤ 1/2 +δ /2 (i) mvr = nh , r == r⎢ δ ⎥ mv m ⎣ ∧R0 ⎦ 1 ⎡ n 2h2 ⎤1– δSolving this for r, we get r = n ⎢ δ ⎥ m ∧ R⎣ 0 ⎦ For n = 1 and substituting the values of constant, we get 1 ⎡ h2 ⎤1– δ r = ⎢1 δ ⎥ m ∧ R⎣ 0 ⎦ 1 2 –68 ⎡ 1.05 ×10 ⎤2.9 r = 8 × 10–11 = 0.08 nm1 = ⎢ –31 –28 +19 ⎥9.1 10 × 2.3 ×10 ×10 ⎦⎣ × (< 0.1 nm) 1 nh ⎛ δ ⎞1– δ hm ∧ R(ii) vn = = nh 0 . For n = 1, v1 = = 1.44 × 106 m/s⎜⎟mr 22 n ⎝ n h ⎠ mr 1 1 2 –19 (iii) K.E. = mv 1 = 9.43 ×10 J=5.9eV 2 ∧P.E. till R0 = – R0 r δ r δ dr ∧R0 ⎡ 1 ⎤P.E. from R0 to r =+∧ R0 ∫ 2+δ =+ ⎢ 1+δ ⎥ Rr –1 – δ ⎣r ⎦ 0R 0 ∧R0 δ ⎡ 11 ⎤ – = – ⎢ 1+δ 1+δ ⎥1+δ ⎣rR0 ⎦ ∧ ⎡ Rδ 1 ⎤ = – 0–⎢ 1+δ ⎥1+δ ⎣rR0 ⎦ ∧ ⎡ R0 δ 11+δ ⎤ PE .. = –– +⎢ 1+δ ⎥1+δ ⎣r RR00 ⎦ ∧ ⎡R –1.9 1.9 ⎤ .. = –PE –⎢ 0 ⎥–0.9 r –0.9 R⎣ 0 ⎦ 2.3 –18 0.9 =×10 [(0.8) –1.9] J = – 17.3 eV 0.9 Total energy is (–17.3 + 5.9) = –11.4 eV. Chapter 13 13.1 13.2 13.3 13.4 13.5 13.6 13.7 13.8 13.9 13.10 13.11 13.12 13.13 13.14 (c) (b) (b) (a) (a) (b) (b) (a), (b) (b), (d) (c), (d) No, the binding energy of HdN dt 13 is greater. N B has shorter mean life as λ is greater for B. Excited electron because energy of electronic energy levels is in the range of eV, only not in MeV. as γ -radiation has energy in MeV. 13.15 2γ photons are produced which move in opposite directions to conserve momentum. 13.16 Protons are positively charged and repel one another electrically. This repulsion becomes so great in nuclei with more than 10 protons or so, that an excess of neutrons which produce only attractive forces, is required for stability. At t = 0, N = N while N = 0. As time increases, N falls off AO B A exponentially, the number of atoms of B increases, becomes maximum and finally decays to zero at ∞ (following exponential decay law). 1 R013.18 t = ln λ R 5760 16 5760 4 = ln = ln 0.693 120.693 3 5760 4 = ×2.303log = 2391.12 years.0.693 3 13.19 To resolve two objects separated by distance d, the wavelength λ of the proving signal must be less than d. Therefore, to detect separate parts inside a nucleon, the electron must have a wavelength less than 10–15 m. λ= h and K ≈ pc ⇒ K ≈ pc = hc p λ 6.63 ×10 34 × 3×10 8 = eV –19 –15 1.6 ×10 ×10 = 109 eV. = 1 GeV. 13.20 (a) 23 1 = 11, N1 = 1211 Na : Z23 23 ∴ Mirror isobar of 11 Na = 12 Mg . 182 (b) Since Z2 > Z1, Mg has greater binding energy than Na. 38 38 38 13.21 S ⎯⎯⎯⎯→ Cl ⎯⎯⎯⎯→ Ar 2.48 h 0.62 h At time t, Let 38S have N1(t ) active nuclei and 38Cl have N2(t) active nuclei. dN1 38 = –λ1N1 = rateof formation of Cl .Also dt dN2 = –λ N +λ N12 11dt – t1But N = Ne λ 10 dN2–λ t = –λ Ne 1–λ N10 22dt λtMultiplying by 2 and rearranging e dt λ t λ t (λ –λ )t2 221e dN +λ N e dt=λ N e dt 2 22 10 Integrating both sides. λ t N λ (λ –λ )t2 01 21N2e = e + C λ – λ21 Since at t = 0, N2 = 0, C=– N0λ1 λ – λ21 λ 2t N0λ1(λ 2–λ1)t∴ Ne = (e –1) 2 λ – λ21 N λ0 1–λ,t –λ tN2 = (e – e 2)λ – λ21 For maximum count, dN 2 = 0 dt ⎛ λ 1 ⎞On solving, t = ⎜ ln ⎟/( λ 1– λ 2)λ⎝ 2 ⎠ 2.48 = ln /(2.48 – 0.62) 0.62 ln 4 2.303 log 4 == 1.86 1.86 = 0.745 s. 13.22 From conservation of energy 22 pn ppE − B = Kn + Kp =+ (1)2m 2m From conservation of momentum E p + p = np (2) c If E = B, the first equation gives pn = pp = 0 and hence the second equation cannot be satisfied, and the process cannot take place. For the process to take place, Let E = B + λ, where λ would be < S , Sn nucleus is more stable than Sb nucleus.pSn pSb(ii) It indicates shell structure of nucleus similar to the shell structure of an atom. This also explains the peaks in BE/ nucleon curve. Chapter 14 14.1 (d) 14.2 (b) 14.3 (b) 14.4 (d) 14.5 (b) 14.6 (c) 14.7 (b) 14.8 (c) 14.9 (a), (c) 14.10 (a), (c) 14.11 (b), (c), (d) 14.12 (b), (c) 14.13 (a), (b), (d) 14.14 (b), (d) 14.15 (a), (c), (d) 14.16 (a), (d) 14.17 The size of dopant atoms should be such as not to distort the pure semiconductor lattice structure and yet easily contribute a charge carrier on forming co-valent bonds with Si or Ge. 14.18 The energy gap for Sn is 0 eV, for C is 5.4 eV, for Si is 1.1 eV and for Ge is 0.7eV, related to their atomic size. 14.19 No, because the voltmeter must have a resistance very high compared to the junction resistance, the latter being nearly infinite. +1V 14.20 0 14.21 (i) 10 × 20 × 30 × 10–3 = 6V (ii) If dc supply voltage is 5V, the output peak will not exceed Vcc =5V. Hence, V0 = 5V. 14.22 No, the extra power required for amplified output is obtained from the DC source. 14.23 (i) ZENER junction diode and solar cell. (ii) Zener breakdown voltage (iii) Q- short circuit current P- open circuit voltage. 14.24 Energy of incident light photon –34 86.6 ×10 × 3 ×10 hv = = 2.06e V–7 –19 6 ×10 ×1.6 ×10 For the incident radiation to be detected by the photodiode, energy of incident radiation photon should be greater than the band gap. This is true only for D2. Therefore, only D2 will detect this radiation. V – V14.25 BB BE If R is increased, I will decrease. Since I = βI , it willIB = . 1B cbR result in decrease in IC i.e decrease in ammeter and voltmeter readings. 1 OR gate gives output according to the truth table. A B C 0 0 0 0 1 1 1 0 1 1 1 1 V014.27 Input Output A A 0 1 1 0 14.28 Elemental semiconductor’s band-gap is such that emissions are in IR region. 14.29 Truth table A B Y 0 0 0 0 1 0 1 0 0 1 1 1 AND Gate 188 P I == 0.2A = 200mA 14.30 Z max VZ V – V 2 R = sZ = = 10 Ω.SI 0.2 Z max 14.31 I3 is zero as the diode in that branch is reverse bised. Resistance in the branch AB and EF are each (125 + 25) Ω = 150 Ω . As AB and EF are identical parallel branches, their effective 150 resistance is = 75 Ω 2 ∴ Net resistance in the circuit = (75 + 25) Ω = 100 Ω . 5∴ Current I1 == 0.05A . 100 As resistances of AB and EF are equal, and I1 = I2 + I3 + I4, I3 = 0 0.05 ∴ I = I == 0.025A 2 4 2 14.32 As Vbe = 0, potential drop across Rb is 10V. 10 ∴ I == 25 µAb 400 ×10 3 Since V = 0, potential drop across R , i.e. IR is 10V. ce ccc 10 –3 ∴ I == 3.33 ×10 = 3.33m A .c 3 ×10 3 Ic 3.33 ×10 –3 2∴β= = = 1.33 ×10 = 133. Ib 25 ×10 –6 14.34 14.35 14.36 From the output characteristics at point Q, VCE = 8V & IC = 4mA V = I RC + V CCC CE V – VCC CE R = cIC 16 – 8 R = –3 = 2K Ω c 4 ×10 Since, V = IR + V BB BB BE 16 – 0.7 R == 510K ΩB 30 ×10 –6 I 4 ×10 –3 Now, β= C == 133 IB 30 ×10 –6 RCVoltage gain = AV = –β RB 2 ×10 3 = –133 × 3510 ×10 = 0.52 Power Gain = A = β× A pVR2 C= –β RB 22 ×10 3 Voltage = (133) × 3 = 69510 ×10 When input voltage is greater than 5V, diode 5V is conducting Time When input is less than 5V, diode is open circuit –(i) In ‘n’ region; number of e is due to As: ne = ND = 1 × 10–6 × 5 × 1028 atoms/m3 ne = 5 × 1022/m3 The minority carriers (hole) is ni (1.5 ×10 ) 2.25 ×10 n == = h 22 22 n 5×10 5 ×10 e nh = 0.45 × 10/m3 Similarly, when Boron is implanted a ‘p’ type is created with holes nh = NA = 200 × 10–6 × 5 × 1028 = 1 × 1025/m3 –This is far greater than e that existed in ‘n’ type wafer on which Boron was diffused. Therefore, minority carriers in created ‘p’ region ni 2 2.25 ×10 32 n == e nh 1×10 25 = 2.25 × 107/m3 (ii) Thus, when reverse biased 0.45 × 1010/m3, holes of ‘n’ region would contribute more to the reverse saturation current than 2.25 × 107/m3 minority e– of p type region. 14.38 14.39 I ≈ I ∴ I (R + R ) + V =12 V CE C CECERE = 9 – RC = 1.2 KΩ 14.40 15.1 15.2 15.3 15.4 15.5 15.6 15.7 15.8 15.9 15.10 15.11 15.12 15.13 ∴ VE = 1.2 V V = V + V =1.7 V BE BE V I = B = 0.085mA 20K 12–1.7 10.3 RB = == 108K Ω IC /β+ 0.085 0.01 + 1.085 I = I + II = β I (1)EC B C B IR + V + IR = V (2)CC CEEE CC RI + V + IR = V (3)B BE EE CC From (3) I ≈ I = β I eC B V – V 11.5 CC BE (R + β R )= V – V , I == mA ECC BEBR +β RE 200 From (2) V–V V–V 2CC CE CC CE R +R === (12 – 3)K Ω= 1.56K ΩCE IC β IB 11.5 RC = 1.56 –1 = 0.56K Ω Chapter 15 (b) (a) (b) (a) (b) (c) (b) (b) (c) (a), (b), (d) (b), (d) (b), (c), (d) (a), (b), (c) 192 15.14 15.15 15.16 15.17 15.18 15.19 15.20 15.21 (b), (d) (i) analog (ii) analog (iii) digital (iv) digital No, signals of frequency greater than 30 MHz will not be reflected by the ionosphere, but will penetrate through the ionosphere. The refractive index increases with increase in frequency which implies that for higher frequency waves, angle of refraction is less, i.e. bending is less. Hence, the condition of total internal relection is atained after travelling larger distance (by 3MHz wave). A + A = 15, A – A = 3 cm cm ∴ 2Ac = 18, 2Am = 12 A 2 m ==∴ m A 3 c = 1MHz LC 1 = 2π×10 6 In AM, the carrier waves instantaneous voltage is varied by modulating waves voltage. On transmission, noise signals can also be added and receiver assumes noise a part of the modulating signal. However in FM, the carriers frequency is changed as per modulating waves instantaneous voltage. This can only be done at the mixing/ modulating stage and not while signal is transmitting in channel. Hence, noise doesn’t effect FM signal. Loss suffered in transmission path = – 2 dB km–1 × 5 km = – 10 dB Total amplifier gain = 10 dB + 20 dB= 30 dB Overall gain of signal = 30 dB – 10 dB = 20 dB ⎛ Po ⎞10log ⎜ P ⎟ = 12 or Po = P i × 102 ⎝ i ⎠ = 1.01 mW × 100 = 101 mW. 15.22 (i) Range = 6 = 16 km2 × 6.4 ×10 ×20 Area covered = 803.84 km2 (ii) Range = 2 × 6.4 ×10 6 × 20 + 2 × 6.4 ×10 6 × 25 = (16 + 17.9) km = 33.9 km Area covered = 3608.52 km2 ∴ Percentage increase in area (3608.52 – 803.84) = ×100 803.84 = 348.9% 2215.23 d = 2( R + h )mT 8RhT = 2(R+hT)2 (Q dm = 22 Rh T ) 4Rh = R2 + h 2 + 2Rh T TT (R – hT)2 = 0 R = h T Since space wave frequency is used, λ << hT, hence only tower height is taken to consideration. In three diamensions, 6 antenna towers of hT = R would do. 15.24 For F1 layer ⎛ 5 ⎞2 )1/2 or N 65 × 106 = 9(N = ⎜ ×10 ⎟ = 3.086 × 1011 m–3 maxmax ⎝ 9 ⎠For F2 layer )1/2 or8 × 106 = 9 (Nmax⎛ 86 ⎞ 11 –3 N = ⎜ ×10 ⎟ = 7.9 ×10 m = 7.9 × 1011 m–3. max ⎝ 9 ⎠15.25 Of ω – ω , ω and ω + ω , only ω + ω or ω – ω contains information. cmc mmc m c m Hence cost can be reduced by transmitting ω + ω , ω – ω , both ω +ω c mc mc m & ω – ω c m 15.26 15.27 x d 15.28 I 1 ⎛ 1 ⎞(i) = , so ln ⎜⎟ = –α x Io 4 ⎝ 4 ⎠ ⎛ ln4 ⎞or ln4 = ax or x = ⎜ ⎟⎝ α ⎠ I (ii) 10log10 = –α x where α is the attunation in dB/km.Io I 1Here = Io 2 ⎛ 1 ⎞or 10log ⎜⎟ = –50 α orlog2 = 5 α ⎝ 2 ⎠ log2 0.3010 or α== = 0.0602 B/km d 55 2x = velocity time 2x = 3 × 108 m/s × 4.04 × 10–3s 12.12 × 105 x= m = 6.06 × 105m = 606 km2 222 22d = x – hs = (606) – (600) = 7236; d = 85.06 km Distance between source and receiver = 2d ≅170 km d = 22 Rh , 2d = d ,4d2 = 8 RhmT m T 2 7236 d = ≈ 0.565 km = 565m.= hT 2× 6400 2R From the figure 100 20 V = = 50V, V = = 10 V. max min2 2(i) Percentage modulation Vmax – Vmin ⎛ 50 –10 ⎞ 40 µ(%) =×100 = ⎜⎟ ×100 = ×100 = 66.67% V + V ⎝ 50 + 10 ⎠ 60 max min V + V 50 +10 max min (ii) Peak carrier voltage = Vc = = = 30V 2 2 2 (iii) Peak information voltage = V m = µV c =× 30 = 20V .3 vt ( )= A(A sin ω t + A sin ω t + A sin ω t)15.29 (a) mm1 mm 2 cc12 ( sin ω t + A sin ω t + A sin 2+ BA ω t)mmm mcc112 2 =( ω t + A sin ω t + A sin ω t)AA sin m 1 m 2 m 2 ccm 1 2 22+B(( A sin ω t + At) + A sin ω t mmm cc1 12 +2Ac (Am sin ω m1t + Am sin ω ct)12 ( sin ω t + A sin ω t + A sin = AA ω t )1 m1 m2 m2 cc 22 22[ sin ω t + A sin ω t + 2AA sin ω t+BA ω t sin mm m1 m 12 m2 m1 m21 m 2 +A2 sin 2 ω t + 2A (A sin ω t sin ω t + A sin ω+ sin ω t]c ccmm cmm c11 22 ( sin ω t + A sin ω t + A sin = AA ω t )m mmm2 cc1 12 22 2222[ sin ω t + A sin ω t + A+BA sin ω t m1 m 1 m 2 m2 cc 2 AA m1 m 2+ [cos( ω – ω )t – cos( ω +ω )] tm 2 m 1 m 1 m 22 2 AA + cm 2 [cos( ω – ω t ω +ω )] ) – cos( t cm 1 cm 12 2 AA + cm1 [cos( ω – ω )t – cos( ω +ω )] t cm cm2 22 ∴ Frequencies present are ω ,ω ,ω m1 m2 c (ω – ω ),( ω +ω )m2 m1 m1 m2 (ω – ω ),( ω +ω )cm1 cm1 (ω – ω 2 ),( ω +ω 2)cm cm (i) Plot of amplitude versus ω is shown in the Figure. ω +ω m1 m2 ω +ω cm1ω m – ω m21 ω – ω cm1 ωω ω – ω ω ω+ωm1 m2 cmc cm22 (ii) As can be seen frequency spectrum is not symmetrical about ω c. Crowding of spectrum is present for ω < ω c. 196 (iii) Adding more modulating signals lead to more crowding in ω < ω c and more chances of mixing of signals. (iv) Increase band-width and ω c to accommodate more signals. This shows that large carrier frequency enables to carry more information (more ω m) and which will in turn increase bandwidth. 1 –3 15.30 f m = 1.5kHz, = 0.7 ×10 s f m 1 –7 f = 20MHz, = 0.5 ×10 s cf c (i) RC = 103 × 10–8 = 10–5 s 11 << RC RC , so this cannot be demodulated. f c

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