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PHYSICS PART TEST-3 13th.pdf
PHYSICS PART TEST-3 13th.pdf
PHYSICS PART TEST-3 13th.pdf
PHYSICS PART TEST-3 13th.pdf
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PHYSICS PART TEST-3 13th.pdf
PHYSICS PART TEST-3 13th.pdf
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PHYSICS PART TEST-3 13th.pdf

  1. Class : XIII (Sterling) Time : 3 hour Max. Marks : 60 INSTRUCTIONS 1. The question paper contain pages and 3-parts. Part-A contains 6 objective question , Part-B contains 2 questions of "Match theColumn" type and Part-C contains 4 subjective type questions.All questions are compulsory. Please ensure that the Question Paper you have received contains all the QUESTIONS and Pages. If you found some mistake like missing questions or pages then contact immediately to the Invigilator. PART-A (i) Q.1 to Q.6 have One or More than one is / are correct alternative(s) and carry 4 marks each. There is NEGATIVE marking. 1 mark will be deducted for eachwrong answer. PART-B (iii) Q.1 to Q.2 are"Match the Column" typewhich mayhave oneor more than onematching options and carry8 marks foreach question. 2 marks will be awarded for each correct match within a question. ThereisNONEGATIVEmarking.Markswillbeawardedonlyifallthecorrectalternativesareselected. PART-C (iv) Q.1 to Q.4 are "Subjective" questions. There is NO NEGATIVE marking. Marks will be awarded onlyifall the correct bubbles are filled in yourOMR sheet. 2. Indicate the correctanswer for each question byfillingappropriate bubble(s) inyour answer sheet. 3. Use onlyHB pencil fordarkeningthe bubble(s). 4. Use ofCalculator, LogTable, SlideRule and Mobile is not allowed. 5. Theanswer(s)ofthequestionsmustbemarked byshadingthecirclesagainst thequestionbydarkHBpencil only. PART TEST-3 PART-B For example if Correct match for (A) is P, Q; for (B) is P, R; for (C) is P and for (D) is S then the correct method for filling thebubbleis P Q R S (A) (B) (C) (D) PART-C Ensure that all columns (4 before decimal and 2 after decimal) are filled. Answer having blank column willbe treated as incorrect. Insert leading zero(s) if required after rounding the result to 2 decimalplaces. e.g. 86 should befilled as 0086.00 . . . . . . . . . . PART-A For example if only 'B' choice is correct then, the correct method for fillingthebubbleis A B C D Forexampleifonly'B& D' choices are correct then, the correct method forfillingthebubblesis A B C D The answer of the question in any other manner (such as putting , cross , or partial shading etc.) will be treated as wrong. PHYSICS
  2. PARTA Select the correct alternative(s) (one or more than alternative may be correct) Q.1 Whichofthefollowingaredimensionlessquantities? (A) 2 CR L (B) 2 2 0 B E    (C) 2 2 B E (D)mLi2 Q.2 to Q4: In the cct shown in figure following events are performed. Event A : switch S1 is closed Event B : switch S2 is closed Event C : switch S3 is closed Event D : switch S4 is closed Q.2 Current drivenbybatteryimmediatelyafter(A + B + C) event : (A) 2 1 R R E  (B) 1 R E (C) zero (D)infinite Q.3 Voltage across inductoris zero : (A)afterlongtimeirrespectiveofanyevents (B) onlywhen event B takes place irrespective of anyother event (C) immediate after event (B + C + D) (D) onlywhen eventAdoes not happen Q.4 Arock ofmass m is droppedinto a pot ofwater, where it fall with a constant acceleration ofg/2.At the same time an equal volume of water of mass m/5 accelerates upward with same acceleration to make room for the rock. The combined mass of water and pot is 2m and entire system rest on a scale. What is the readingofscale : (A) mg 5 13 (B)mg (C) mg 3 5 (D) none Q.5 A tank is filled upto a height h with a liquid and is placed on a platform ofheighth fromtheground.To getmaximumrange xm a small hole is punched at adistance of y from the free surfaceoftheliquid.Then: (A) xm = 2h (B) xm = 1.5 h (C) y = h (D) y = 0.75 h Q.6 Surface tension of a soap is T. How much work is needed to blow a soap bubble of Radius R ? (A) 2R2 T (B) 4R2 T (C) 8R2 T (D) 16R2 T
  3. PART B Match the column Q.1 ApointchargeqismovinginacircleofradiusRwithangularfrequency.Matchthefollowingforgiven data : Table - 1 Table - 2 (A)Equivalentcurrent (P) Proportional to (q/R) (B)Equivalentmagneticmoment (Q) Proportional to (q) (C)Magneticfield at centre (R) Proportional to (qR2 ) (D)Torque actingon charge about centre of circle (S) None Q.2 Aseries L–C circuit is connected toanAC source of220Vand 50 Hz as shown in figure.If the reading three idealA.C. voltmeters are V1 ,V2 and V3 are 65V, 415V and 204V respectively : Column - I Column - II (A)rms valueofcurrentincircuit (P) 1 S.I. unit (B)The valueofinductance‘L’(approximately) (Q) 0.65 S.I. unit (C) The value of capacitance (C)(approximately)(R) 10×10–6 S.I. unit (D) The value of C (for the same L) required to (S) 5×10–6 S.I. unit produce resonance PART C Q.1 Mercuryiscontained in U-tube ofuniform square cross-section side a = 4mm. ElectrodesAand D are sealed inside as shown in the fig.An electric current I= 10Apasses between the electrodes fromAto D and magnetic field B = 5.44 wb/m2 is appliedacrossthehorizontal arm perpendicularintothe plane of the figure ? Here densityof mercury= 13.6 × 103 kg/m3 , g = 10 m/s2 , Nelgect surface tension. Calculate the difference (in mm)inmercurylevel inlimbsofUtube. Q.2 Thefigureshows acopper rod(PQ) movingwith velocityv parallel to a longstraight current carryingwire havingcurrent I=100A.Calculate the current (in mA) in copper rod PQ. Here v = 5 m/s., a = 1 cm, b = 100 cm and R = 9.2×10–3 . Q.3 An electron of kinetic energy100 eV collides with stationaryhelium ion (He+ ) in its ground state and excites it to a higher level.After collision He+ ion emits two photons in succession with wavelengths 1080Å and 300Å. Find the kinetic energy (in e.V.) of electron after collision. Neglect the recoil speed of He+ ions and hC=12420 eV–Å. Q.4 A metal disc of radius R =6 cm is mounted on a frictionless axle. The current canflow throughtheaxleout alongthediscto asliding contact ofrimofthe disc.Auniform magneticfield B=2Tis parallel to the axis of the disc. When the current is 3A, the disc rotates with constant angularvelocity.What is thefrictional force(inNewton)at therim betweenthestationaryelectrical contact andtherotating rim?
  4. SOLUTION PART A Select the correct alternative(s) (one or more than alternative may be correct) Q.1 [Ans:A,B] [Sol: CR =  (time const)  = 2 CR L =   = 1 R L = , 0 0 1   = v,  = 0 E0 × 2 2 B E = 1 ] Q.2 [Ans:A] [Sol: ImmediatelyafterA+ B + C current through Lis zero & charge on C1 is zero from KvL i = 2 1 R R   ] Q.3 [Ans:A] [Sol: inductor is ideal, at steady statecorrentwill beomeconstant ] Q.4 [Ans:A] [Sol. Wwater + Wpot + Wrock N = (Wwater + Wrock + Wpot )g = Mhal aCM here aCM = M a W a W water sigh rock rock  mg 5 13 N  ] Q.5 [Ans:A] [Sol: We know Rmax = H at y = 2 H Here H  h + h = 2h There fore maximum range will be 2h at y= h, i : e hole should be at bottom. ]
  5. Q.6 [Ans:C] [Sol:  W = (2A) T = 2 × 4R2 × T (A bubblehas two faces)] PART B Q.1 [Ans: (A)  Q; (B)  R; (C)  P; (D)  S] [Sol: i = dt dq =   2 × q i × q magnetic moment M = IA = (q)R2 magnetic field at centre B = R 2 i 0  = R 2 0  × q   2 B × R q Ans: (A)  Q; (B)  R; (C)  P; (D)  S] Q.2 Ans: (A) - Q; (B) - P; (C) - S; (D) - R PART C Q.1 [Ans:100mm] [Sol.AccordingtoFleming’s left handrule theentirebodyof mercuryexperiences magneticforceFm =BIa.To balance this forcea difference inthe mercurylevel’s is setup.Let h be thedifferencein level 2 a ) gh ( a B     m 1 . 0 10 1 10 4 10 10 6 . 13 10 44 . 5 ga BI h 3 3            = 100 mm.] Q.2 [Ans:50mA] [Sol. Motional emf develop across rod PQ        b x dx Bv d dx v x I 2 a x 0      a b n 2 Iv 0      100 n 2 5 100 10 4 7         V 10 6 . 4 4    induced current inrod A 05 . 0 20 1 10 2 . 9 10 6 . 4 R i 3 4 ind          mA 50  ]
  6. Q.3 [Ans. 47.1 e.V.] [Sol. 1 = 1080Å  1080 12420 hC E 1 ' 1    = 11.50 eV 1 = 300 Å  300 12420 E' 2  = 41.4 eV 9 . 52 E E ' 2 ' 1   eV 9 . 52 E E E E 1 n ' 2 ' 1     eV From emfyconsv. n ' 1 1 1 E K E K     ) E E ( K K 1 n 1 ' 1    = 100 – 52.9 V . e 1 . 47 K' 1  ] Q.4 [Ans: 18N] [Sol. 0     0 B f             B Id r f R f            0 f ) k ( dr Br I ) k̂ ( sin Rf ) k ( 2 B IR ) k̂ ( Rf 2 r     Ans. 2 IRB fr  = 0.18 N]
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