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Class 10 ISC Prelims 2017 : Physics (Smt. Lilavatibai Podar High School (LPHS), Mumbai)

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Harshitha
St. Ann's High School ICSE / ISC, Secunderabad
7th to 12th science stream

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XII PHYSICS THEORY PAPER Solutions (v) Q.1. A. M.C.Q. : (i) (A) q (ii) (D) All of these (iii) (A) Speed remains constant (D) n = 2 to n = 1 (v) (C) Fringe width decreases (ii) Zero When no current is drawn from a cell by the electric circuit OR Open circuit (iii) e=L di dt 2=L (10 0) 0.40 L= (iv) a1 + a2 2 = a1 a2 1 a1 3 = a2 1 0.40 2 = 0.08 H 10 I1 1 a21 9 = = = I2 2 a22 1 High: retentivity coercivity,susceptibility (x) (i) (ii) (iii) (xi) c or 90 perpendicular to each other 2 (vii) 2D (viii) A telescope is said to be in normal adjustment when it forms the final image of a distance object at infinity. 2 Imax (a1 + a2) 4 (ix) = = Imin (a1 a2)2 1 (vi) (iv) B. (i) Interference Due to superposition of light waves from two coherent sources. Fringes may or may not be of same width. All bright fringes are of same intensity. = Pavg = Irms Vrms cos R = 1.5 200 Z 4 = 300 5 = 240 W Diffraction (i) Due to interference of secondary wavelets coming from different points of same wavefront. (ii) Fringes are never of same width. (iii) Intensity of successive bright fringes goes on decreasing. E 42 eV = c 3 108 SECTION A Q.2. 42 1.6 10 19 = 8 3 10 27 = 22.4 10 (a) kg m/s or Js/m (xii) Electromagnetic waves are produced by accelerated charged particles while matter waves are associated with moving particles irrespective of charge on it.OR EM waves do not medium while matter waves need. (xiii) Lyman series (xiv) Emitter base : Forward biased : electric field intensity due to (+q) E2 : electric field intensity due to ( q) Pt. O : centre of dipole AB Pt. P : point where electric field intensity is to be calculated Collector base : Reverse biased (xv) E1 It indicates the stability of the nucleus. 1 Theory 2015 Second Term Solutions The components E1 sin and E2 sin are equal and opposite, so cancel out each other. While the components E1 cos and E2 cos are equal and in same direction and so get added up. E = E1 cos + E1 cos E = 2E1 cos KVL to loop BCDEB : 4I2 8 + 1I2 + 4I2 + 8(I1 + I2) = 0 4I2 8 + 1I2 + 4I2 + 8I1 + 8I2 = 0 8I1 + 17I2 = 8 Eqn. (1) 4 8I1 4I 2 8 8I1 17 I 2 9 13I 2 0 ( E1 = E2) Now E1 = 1 q 4 0 r2 + 2 (away from +q) E2 = 1 q 4 0 r2 + 2 (towards q) I2 = 0 A Also cos = 2 8I1 + 4(0) = 8 2 r + I1 = 1 A From (1), E =2 ... (2) VBE = 8(I1 + I2) 1 q 4 0 r2 + 2 (r2 + 2)1/2 = 1 (2q) 2 2 3/2 4 0 (r + ) = 1 P [ P = 2q ] 4 0 (r2 + 2)3/2 (b) Let = 8(1 + 0) =8V C1 = C2 =C3 = C Cp = C1 + C2 + C3 = 3C 1 1 1 1 3 = + + = Cs C1 C2 C3 C For short dipole (r >> 2 ) 1 P kP E= = 4 0 r3 r3 Cs = C 3 Cp 3C 9C = = =9:1 Cs C/3 C Direction of E is antiparallel to dipole axis. (b) Q.4. (a) The average (or constant) velocity of electrons attained when a potential difference is applied across the ends of the conductor. No. of electrons N n : Electron density = = Volume V i : current through conductor Q : total charge e : electron charge Q i= (1) t Acc. to quantization of charge, Q = Ne = (nAvdt) e U = UAB + UBC + UAC = 1 1 [Q Q + Q2Q3 + Q1Q3] 4 0 r 1 2 1 12 10 [(25) (50) + 7.5 (50) (100) + (25) (100)] = 9 10 9 = 9 10 3 [1250 + 5000 + 2500] 7.5 Q = neAvd t i = neAvd U = 10.5 J Q.3. (a) KVL to loop ABEFA : 3I1 8(I1 I 2 ) 3I1 2I1 16 10 3I1 8I1 8I 2 3I1 2I1 16 0 16I1 8I 2 16 ( 8) 2I1 + I2 = 2 i = neAvd A j = nevd j = current density vd = drift velocity ... (1) 2 Theory 2015 Second Term Solutions r = 1 + m (b) (i) (a) Principle : When electric current flows in a coil placed in an uniform radial magnetic field, a deflecting torque acts upon the coil whose magnitude depends on the strength of current. 1.000022 = 1 + m m = 0.000022 (ii) Curie s law : Magnetization (M) of a paramagnetic substance is directly proportional to magnetic intensity (H) of magnetizing field and inversely proportional to Kelvin temperature (T). H M = C C : Curie constant T OR Magnetic susceptibility is inversely proportional to Kelvin temperature Q.5. (a) Deflecting torque : = NiAB sin : angle between magnetic field and normal to coil Maximum torque : d = NiAB Restoring torque : R : angle of twist (i) Consider a rectangular loop of conducting wire PQRS partly placed in an uniform magnetic field of induction (B). R = C (ii) Let be the length of side PS and x be the length of loop within the field. C : torsional constant In equilibrium position, A = x = area of loop Magnetic flux, = BA = B x d = R Differentiate w.r.t. time (t) NiAB = C d d dx = (B x) = B dt dt dt C i= NAB d = B v dt But e = (b) ig = i d dt (b) e = B v 50 div = 10 mA 5 div/mA ig G S= i ig 3 10 10 40 2= i 10 10 3 i = 0.21 A Q.6. 3 Theory 2015 Second Term Solutions Time constant ( ) of L-R circuit : SECTION B (i) It is the time in which current goes from zero to 0.632 of its steady value. Q.8. (ii) It is the time in which current decays from maximum to 0.368 of the maximum value. (a) (iii) It is the ratio of coefficient of induction to resistance (R). = L R Q.7. (a) E= 2 2 VL + VR = 2002 + 1502 E = 250 V i=5A Consider two coherent sources S1 and S2 separated by distance d (slit width). E 250 Z= = = 50 i 5 d d BP = x + , AP = x 2 2 In S2PB, Pythagoras theorem S2P2 = BP2 + S2B2 Z = 50 VL XL i VL 200 4 tan = = = = = R VR VR 150 3 i 2 d S2P2 = x + + D2 2 In S1AP : S1P2 = AP2 + S1A2 4 = tan = 53 7 3 1 d 2 S1P2 = x + D2 2 (b) Magnetic field at point P due to infinitely long conductor, B= Consider, S2P2 S1P2 d 2 d 2 = x + + D2 x D2 2 2 d d2 d d2 2 = x + 2x + x2 + 2x 2 4 2 4 S2P2 S1P2 = 2xd (S2P S1P) (S2P + S1P) = 2xd 2xd S2P S1P = D+D 0 2i 4 R B = 10 5 NA 1m 1 B is perpendicular to the plane of page and directed downwards. Path difference = F = evB sin 2xd xd = 2D D Case I : Point P is bright (constructive interference) P D = even multiple of 2 xd = (2n) D 2 = 1.6 10 18 N By Fleming s left hand rule, F on electron is perpendicular to the plane of page and directed away from the wire. x = (2n) 1st bright D 2d n = 0,1,2 x=0 2 D 2 bright n=1 x= 2d 4 D 3rd bright n=2 x= 2d 2 D D Fringe width = 0= 2d d nd 4 n=0 Theory 2015 Second Term Solutions = = 4 D 2 D D = 2d 2d d The focal length of each half is 2f (doubled) and power is P/2. (b) Given : A = 60 , m = 30 D d To find : v = ? Case II : If point P is dark P D = odd multiple of 2 xd = (2n 1) n = 1, 2, D 2 x = (2n 1) Sol. : Prism formula A + m sin 2 = A sin 2 D 2d 60 + 30 sin 2 sin 45 = = = 1.414 60 sin 30 sin 2 1 D 2d 3 D 2nd dark point n=2 x= 2d 3 D 1 D D Fringe width = = 2d 2d d 1st dark point n=1 x= Absolute R.I. : c = v D = d 1.414 = 3 3 5890 10 10 sin = = 2e 2 0.25 10 3 (b) sin = 0.0035 sin is very small = 0.0035 rad Q.9. (a) For equiconvex lens, R1 = R2 = R 1 1 1 ( 1) f R R 1 2 1 1 1 2 ( 1) = ( 1) f R R R f= R 2( 1) On cutting in thickness, each half becomes a planoconvex lens (R1 = , R2 = R). Focal length (f ) is 1 1 1 1 = ( 1) = f R R R f = = 2f 1 5 3 108 v v = 2.122 108 m/s Theory 2015 Second Term Solutions Q.10. (a) Expression for magnifying power : Image formed by compound microscope : angle subtended by object at the eye when placed at DDV : angle subtended by image at the eye fo : focal length of objective fe : focal length of eyepiece uo : distance of AB from objective ue : distance of A1B1 from eyepiece vo : distance of A1B1 from objective ve : distance of A2B2 from eyepiece Since and are small angles tan = = AB D tan = = A2B2 A1B1 = (2) ve ue (1) From equations (2) and (2), we have M.P. = = 6 A1B1 AB = / ue D A1B1 D AB ue Theory 2015 Second Term Solutions M.P. = vo D uo ue (b) (3) Equation (3) can be written as vo D M.P. = (4) uo ue For eyepiece, 1 1 1 = fe ve ue D D D = fe ve ue SECTION C D D D = + ue ve f e Q.12. (a) D D D = + (considering sign conventions) ue ve fe From equation (4), vo D D M.P. = + uo ve fe D vo Here ve = D, So M.P. = 1 + u o fe (b) = (b) h 2mE When K.E. is made 4E, then wavelength is . h = = = 2m 4E 2 2mE 2 OA = OB = r 1 1 3 = = = sin C 4/3 4 (c) E1 = 2W0, E2 = 5W0 (Ek)1 = E1 W0 = 2W0 W0 = W0 In OAS, (Ek)2 = E2 W0 = 5W0 W0 = 4W0 OA r = OS h r = h tan C 3 =7 = 3 7 cm 7 tan C = (Ek)1 (Ek)2 = W0 1 = 4W0 4 1 2 mV1 2 1 = 1 4 2 mV2 2 Area of surface through which light passes is = r2 22 2 = ( 3 7) 7 V1 1 = =1:2 V2 2 = 198 cm2 Q.13. Q.11. (a) Bohr s postulates : Postulate 1 : The centripetal force required for the circular motion of electrons is provided by the electrostatic force of attraction between the positively charged nucleus and negatively charged electrons. (a) (i) Plane of polarization : The plane containing the direction of propagation of light and perpendicular to the plane of vibration is called plane of polarization. (ii) ip + r = 90 2 mv 1 (+e) ( e) = 2 r 4 0 r ip = 90 32 = 58 = tan ip = tan 58 = 1.6 Postulate 2 : Electrons can revolve only in certain specified orbits called as stationary or stable orbits in 7 Theory 2015 Second Term Solutions which the angular momentum of electrons is an h integral multiple of . 2 nh L = mvr = 2 n : principal quantum number Postulate 3 : When an electron jumps from a higher energy orbit to a lower energy orbit, it emits out a photon whose energy is equal to the difference in the energy levels of two states. E = h = En En 2 1 8 Theory 2015 Second Term Solutions 1 u = 1.6605 10 27 kg According to Bohr s first postulate, 2 2 mv 1 e = 2 r 4 0 r 27 0.0013 10 = 0.000782 u 1.6605 10 27 Energy released = 0.000782 931 MeV = 0.73 MeV (b) Half life = T = 140 days N 1 n 15 1 1 4 = =1 = = N0 2 16 16 2 m = 2 1 e 2 K.E. = mv = 2 8 0r P.E. = (charge of electron) (electric potential) 1 e = ( e) 4 0 r n = no. of half lives = 4 Time of decay t n= = Half life T t 4= t = 560 days 140 e 4 0r 2 P.E. = T.E. = K.E. + P.E. 2 2 e e = + 8 0r 4 0r Q.15. e2 1 = 1 4 0r 2 (a) e2 8 0r 0h2n2 = But r = me2 e 2 2 0h n 2 En = En = me2 me4 2 8 0h2n2 me4 2 8 0h2 (b) 1 = 8 0 En = = constant = 13.6 eV 13.6 eV n2 hc hc ; 2 = ; V1 = 20 kV; V2 = 30 kV eV1 eV2 Fractional change in minimum wavelength 1 2 V2 V1 = = V2 1 Vi : a.c. input voltage P1, P2 : primary coils of a transformer S1, S2 : secondary coils of a transformer RL : load resistance 30 kV 20 kV 30 kV = 0.333 = Vo : output voltage (b) NAND GATE (AND + NOT) : Q.14. (a) 1 0n neutron 1 H1 proton 0 1 electron antineutrino A B A B Y A B 0 0 0 1 0 1 0 1 1 0 0 1 1 1 1 0 Output is high when at least one of the inputs is low or else it is low. Mass defect m = [mass of proton (1H1 + mass of electron ( 1 0)] [mass of neutron] = [(1.6725 10 27) + (0.009 10 27)] [1.6747 10 27] = 0.0013 10 27 m = 0.0013 10 27 kg 9

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