QUARTERLY KEY ANSWERS FOR PHYSICS (06-07) Note :. For answer in part II, III and IV like reasoning, explaining, narrating, describing and listing the points students may write in their own words but without changing the concepts and without skipping any point. Answers written only in black (or) blue should be evaluated PART I Q.No ANSWER Q.No ANSWER (a) 6 (a) 0.4 H (d) Zero 7 (d) Brushes (c) H = VIt 8 (b) 0 A 4 (d) Electric potential 9 (b) Power transmitted in a direction perpendicular to both the fields (d) 4 pf 0 (a) Contracts 6 (a) C N - m - (a) Pure line spectrum 7 (b) n : No answer (correct ans : ) 8 (a) Drift velocity (a) Plane 9 (a) Decreases 4 (c) Sommerfeld 0 (c) The velocity of the particle (c) :4:9 (b) 0 C 6 (c) ν Z (d) Infinite resistance 7 (b) Absorbs green light (a).7 A 8 (c) Opposite to the direction of motion of oil droplet 4 (c) Unchanged 9 (a) K line (b) Room heater 0 (c) Capacitor PART II Note :. For all problem type questions correct answer without unit reduce half mark. For wrong answer with correct unit do not award mark for unit Electric polarization: The alignment of the dipole moments of the permanent or induced dipoles in the direction of applied electric field is called polarisation or electric polarisation. p E (or) p = αe Applications of a capacitor: i) They are used in the ignition system of automobile engines to eliminate sparking. ii) They are used to reduce voltage fluctuations in power supplies and to increase the efficiency of power transmission. iii) Capacitors are used to generate electromagnetic oscillations and in tuning the radio circuits. J.SHANMUGAVELU [P.G. T. in Physics] Ph. No:99467 http://www.padasalai.net/06/09/th-quarterly-exam-key-answer-06-7.html
τ = pe sin θ (or) = =.4 0 0. 0 4 ( Substitution ) = 8. 0 6 N m. ( correct answer alone ) 4 Electrochemical equivalent : The electrochemical equivalent of a substance is defined as the mass of substance liberated in electrolysis when one coulomb charge is passed through the electrolyte. Its unit is kg C Voltage law: The algebraic sum of the products of resistance and current in each part of any closed circuit is equal to the algebraic sum of the emf s in that closed circuit. ( if closed circuit is not mentioned reduce mark ) 6 Emf Potential difference. The difference of potentials between the two terminals of a cell in an open circuit is called the electromotive force (emf) of a cell. The difference in potentials between any two points in a closed circuit is called potential difference.. The emf is independent of external resistance of the circuit, Potential difference is proportional to the resistance between any two points. 7 I = =. ( Substitution ) t = 80 s ( correct answer alone ) ( or ) I = ( or ) t = 8 Seebeck effect : a circuit consisting of two dissimilar metals like iron and copper, an emf is developed when the junctions are maintained at different temperatures. 9 Ampere s circuital law: The line integral!"#.$% """# for a closed curve is equal to μ 0 times the net current I o through the area bounded by the curve. (OR) Mere writing!"#.$% """# = μ 0 I 0 of the notations 40 Nichrome is used as heating element : i) It has high specific resistance ii) It has high melting point iii) It is not easily oxidized 4 methods of producing induced emf : The induced emf can be produced by changing i) The magnetic induction (B) ii) Area enclosed by the coil (A) iii) The orientation of the coil (&) with respect to the magnetic field. If the word changing is not mentioned no marks J.SHANMUGAVELU [P.G. T. in Physics] Ph. No:99467 http://www.padasalai.net/06/09/th-quarterly-exam-key-answer-06-7.html
4 Q factor: The Q factor of a series resonant circuit is defined as the ratio of the voltage across a coil or capacitor to the applied voltage. ( OR ) '()*+,- +/0( (0 i.e. Q = +44)-6 '()*+,- 4 Induced emf e = - M < < 7= 8 9: ; ( OR) M = - - = >? M = - B C D.C ( F H ) =?? A ( OR) M = - - >@ A @ ( Substitution ) M = 6. 0 H ( OR) M = 6. mh. ( ans, unit ) 44 Conditions for sustained interference i) The two sources should be coherent ii) Two sources should be very narrow iii) The sources should lie very close to each other to form distinct and broad fringes. 4 Optic axis : Inside a double refracting crystal there is a particular direction in which both the rays travel with same velocity. This direction is called optic axis. 46 S = J (OR) C = J K L K M = ( Substitution ) = 0.08 g / cc 47 Applications of Moseley s law: i) Any discrepancy in the order of the elements in the periodic table can be removed by this law by arranging the elements according to the atomic numbers and not according to the atomic weights. ii) Led to the discovery of new elements like hafnium (7), technetium (4), rhenium (7) etc. iii) helpful in determining the atomic number of rare earths, thereby fixing their position in the periodic table. 48 i) Is coherent, with the waves, all exactly in phase with one another, ii) Does not diverge at all and iii) Is extremely intense. ( any other correct answer ) 49 Soft X rays Hard X rays. Wavelength is 4 Å or above Wavelength is in the order of Å. Have lesser frequency and lesser energy Have high frequency and high energy. Have low penetrating power Have high penetrating power 4. They are produced at low potential They are produced at high potential ( any three differences ) x J.SHANMUGAVELU [P.G. T. in Physics] Ph. No:99467 http://www.padasalai.net/06/09/th-quarterly-exam-key-answer-06-7.html
0 For longest wavelength, d sin θ= nλ ( OR ) d (sin θ) max = λ max λ max =.8 0 0 / ( Substitution ) λ max =.64 0 0 m ( OR ).64 A ( ans, unit ) PART III Electric potential energy of an electric dipole in an electric field. τ = pe sin θ U = pe cos θ θ = 0 ; U = pe Principle of a capacitor C = q/v The capacitance depends on the geometry of the conductors and nature of the medium. Wheatstone s bridge Applying Kirchoff s current law to junction B, I I g I = 0 junction D I + I g I 4 = 0 Applying Kirchoff s voltage law to closed path ABDA I P + I g G I R = 0 Applying Kirchoff s voltage law to closed path ABCDA I P + I Q I 4 S I R = 0 For balancing condition I g = 0. N O = P M 4 A l =.00 A l ( or ) A =.00A 4 J.SHANMUGAVELU [P.G. T. in Physics] Ph. No:99467 http://www.padasalai.net/06/09/th-quarterly-exam-key-answer-06-7.html
R = QK R ; R = QK R Calculation Change in resistance = (.00 ) = 0.00 Change in resistance in percentage = 0.00 00 = 0.% ( OR ) S T = K K x = 9.0Ω ρ = UV T K calculation ρ =.08 0-6 Ωm ( ans, unit ) Thermopile Principle Construction working 6 a galvanometer is converted into an ammeter by connecting a low resistance in parallel with it. Ig. G = (I- Ig)S ( OR ) Ig. G = IsS S= G. W X WYW X 7 Energy losses in a transformer i) Hysteresis loss ii) Copper loss iii) Eddy current loss iv) Flux loss ( heading only- mark) J.SHANMUGAVELU [P.G. T. in Physics] Ph. No:99467 http://www.padasalai.net/06/09/th-quarterly-exam-key-answer-06-7.html
8 Average Power in an ac circuit e = Eo sin ωt i = Io sin (ωt + φ) Upto P +' = E 0\ I 0\ Cos ^ 9 sin & = Nm _ sin & S = 0 7070 0-0 & S = 4 sin &` = 0 000 0-0 &` = 0 a 8 - a b = 60 Applications of Polaroid (Any five applications ) 6 Hydrogen spectral series of five series Formula 6 c = R d f x x = R d c f = RdC f = R d c Hf = Rd Calculation _ = 486 f x PART II 6 J.SHANMUGAVELU [P.G. T. in Physics] Ph. No:99467 http://www.padasalai.net/06/09/th-quarterly-exam-key-answer-06-7.html
6 Electric potential due to a dipole: Electric Potential due to +q charge = Electric Potential due to - q charge = HUk V = Y A HUk V Upto V = HUk = V V A V = V =+ < V cos&a 0 Upto V = V = < V cos&a V = q /(J HUk V Special cases :. Θ = 0 ; V =. Θ = 80 ; V = u HUk V u HUk V. Θ = 90 ; V = 0 64 Electric field due to uniformly charged spherical shell At a point outside the shell. Upto E = w Hπε 0 At a point on the surface. E = w Hπε P At a point inside the shell. Upto E = 0 Electric field due to a uniformly charged thin shell is zero at all points inside the shell. 6 Expression for the magnetic induction, at a point along the axis of a circular coil carrying current 0 7 J.SHANMUGAVELU [P.G. T. in Physics] Ph. No:99467 http://www.padasalai.net/06/09/th-quarterly-exam-key-answer-06-7.html
$! """""# = x y <K z{j HU V & = 90 $! """""# = x y <K HU V for two components of db Total magnetic induction due to entire coil B = $! }~ B = B = x y $% HU V B = x y HU V ƒ ( ) 0 66 Eddy currents applications, methods to reduce Eddy current definition of any three applications x ( For mere mentioning the headings mark each ) Eddy current can be minimised by using thin laminated sheets instead of solid metal. holes drilled in the plates 6 0 67 R, L, C series circuit Phasor diagram e = Eo sin ωt V R = I R V L = I X L Vc= I Xc Upto ɸ = tan - = T YT A S instantaneous current = I 0 sin (ˆ ±ɸ) 0 8 J.SHANMUGAVELU [P.G. T. in Physics] Ph. No:99467 http://www.padasalai.net/06/09/th-quarterly-exam-key-answer-06-7.html
Ž 68 Raman effect Raman effect definition s for three types of lines x Raman Shift or Raman frequency Δν = ν ο ν s. The Raman shift does not depend upon the frequency of the incident light but it is the characteristic of the substance producing Raman effect. The intensity of Stoke s line is always greater than Anti stoke s Line. Three energy level diagrams x 69 Expression for bandwidth using Young s double slit experiment 0 (Reduce mark - if the Waves arrive in phase or out of phase depending upon the path difference between two waves is not mentioned ) Upto path difference δ= Œ6 0 condition for bright fringe (nλ) for bright band x = nλ œ + 6 Žcondition for dark fringe (n ) ž for dark band x = 6 (n ) ž Ÿ + definition for Band width (β) upto β= 6 λ 70 J.J Thomson method for determine the specific charge of an electron. : Principle Construction Determination of velocity v = 0 Upto y = - ) \ 9 J.SHANMUGAVELU [P.G. T. in Physics] Ph. No:99467 http://www.padasalai.net/06/09/th-quarterly-exam-key-answer-06-7.html
y = k y - = \ ) Prepared by J. Shanmugavelu [P.G. Assist in Physics] Lions Mat. Hr. Sec. School Jeyam Tuition centre Paramakudi ( Maninagar) Ph. No: 99467 Email: shaam.breeze@gmail.com 0 J.SHANMUGAVELU [P.G. T. in Physics] Ph. No:99467 http://www.padasalai.net/06/09/th-quarterly-exam-key-answer-06-7.html