Anúncio

Anúncio

Próximos SlideShares

PHYSICS PART TEST-2 13th.pdf

Carregando em ... 3

29 de Mar de 2023•0 gostou## 0 gostaram

•3 visualizações## visualizações

Seja o primeiro a gostar disto

mostrar mais

Vistos totais

0

No Slideshare

0

De incorporações

0

Número de incorporações

0

Baixar para ler offline

Denunciar

Educação

CLASS : XII (NUCLEUS) PART TEST- 2 INSTRUCTIONS 1. The question paper contain pages and 3-parts. Part-A contains 6 objective question , Part-B contains 2 questions of "Match the Column" 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 each wrong answer. PART-B (iii) Q.1 to Q.2 are "Match the Column" type which may have one or more than one matching options and carry 8 marks for each question. 2 marks will be awarded for each correct match within a question. There is NO NEGATIVE marking. Marks will be awarded onlyif all the correct alternatives are selected. PART-C (iv) Q.1 to Q.4 are "Subjective" questions. There is NO NEGATIVE marking. Marks will be awarded only if all the correct bubbles are filled in your OMR sheet. 2. Indicate the correct answer for each question by filling appropriate bubble(s) in your answer sheet. 3. Use only HB pencil for darkening the bubble(s). 4. Use of Calculator, Log Table, Slide Rule and Mobile is not allowed. 5. The answer(s) of the questions must be marked by shading the circles against the question by dark HB pencil only. 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 the bubble is 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 will be treated as incorrect. Insert leading zero(s) if required after rounding the result to 2 decimal places. e.g. 86 should be filled as 0086.00 Class - XII Physics Part Test - 2 PART A Select the correct alternative(s) (choose one or more than one) Q.1 A uniformly charged thin non conducting shell of radius R is charged with Q units of charge. The escape velocity of a particle of charge –q and mass m from surface of conductor is (A) [Sol: Vescape = qE (B*) (C) 1 Q (D) geff = m where E = 40 R 2 Vescape = = (B) ] Q.2 A test tube of length l is immersed invertically into a vessel containing mercuryof density (m). It is immersed in such a way so that its closed end just touches the mercury level. Mercury was found to rise to a height x in the tube as shown. The pressure at A is (Here Atmospheric pressure is equal to l length of mercury): (A) (l – x) mg (B) xmg (C*) (2l – x) mg (D) (l – 2x) mg [Sol: Patm = P0 = lg P = P0 + lg – xg = (2l-x)g (C) ] Q.3 An adiabatic container is divided into two equal parts, with an ideal gas at pressure P and temperature

STUDY INNOVATIONSSeguir

Educator em Study InnovationsAnúncio

Anúncio

Anúncio

- Class : XII (NUCLEUS) 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-2 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.
- Class - XII Physics Part Test - 2 PARTA Select the correct alternative(s) (choose one or more than one) Q.1 Auniformlychargedthinnonconductingshellofradius RischargedwithQunitsofcharge.Theescape velocityof aparticle of charge –qand mass m from surfaceof conductor is (A) R 4 Qq 0 (B*) mR 2 Qq 0 (C) mR Qq 0 (D) mR 4 Qq 0 [Sol: Vescape = R · g 2 eff geff = m qE where E = 2 0 R Q 4 1 Vescape = R · R 4 Q · m q 2 2 0 = mR 2 Qq 0 (B) ] Q.2 Atesttubeoflengthlisimmersedinverticallyintoavesselcontainingmercuryof density(m).Itis immersedinsuchawayso thatits closedendjust touches the mercurylevel.Mercurywasfoundtorisetoaheightxinthetubeasshown.The pressureatAis (HereAtmospheric pressure is equal to l length of mercury): (A) (l – x) mg (B)xmg (C*) (2l – x) mg (D) (l – 2x) mg [Sol: Patm = P0 = lg P = P0 + lg – xg = (2l-x)g (C) ] Q.3 Anadiabatic containeris dividedintotwoequal parts, withanidealgas atpressurePand temperatureT at oneside and vacuum onthe other side. Onmaking a hole in the separating wall (A) work done bygas will be nRT (B*)changeininternal energyofgas will bezero (C*) internal energydepends on temperature of gas (D*) work done bygas is zero [Sol: Thethermodynamicprocess will befree-expansionin nature. Hence dw = 0 since d = 0, dU also is zero (B), (C), (D) ] Q.4 Figure shows a network of capacitors and resistances. Potentials ofsomeof thepoints are given. Then (A*) potential at M = 2.6 V (B*) Potential at N = –1.6 V (C*) Charge across 2f capacitor is 8.4 C (D) Charge across 2f capacitor is 4.2 C
- [Sol: 2i1 + 3(i1 + i2 – i3) + 4(i1 + i2) – 10 = 6 i2 + 2i3 = 3 2(i1 + i2 – i3) – 6i3 = 0 solving(A,B, C) ] Q.5 As aliquidof densityflows throughthe tubeshownbelow in astreamlines way (A) volume rate of flow reduces from a1 to a2 (B) volume rate of flow increases from a1 to a2 (C*) h is a measure of pressure difference atAand B (D*) pressure atAhas to be more than that at B by hg [Sol: Rateofflowremains constant C, D ] Q.6 Twolargeblackplanesurfaces aremaintained at constant temperatureT1 &T2 (T1 >T2) & placed in vacuume. Two thin black plates are placed between the twosurfacesandisparalleltothese.Aftersometimesteadyconditionareobtained. By what factor () is the steady heat flow reduced due to presence of black plates? (A) = 2 1 (B*) = 3 1 (C) = 1 (D) = 0 [Sol: Q = (T1 4 – T3 4) = (T3 4 – T4 4) = (T4 4 – T2 4) 3Q = (T1 4 – T2 4) = Q0 Q = 3 Q0 ] PART B Match the column Q.1 In the circuit shown in figure, there are three identical capacitor C1,C2, C3 each having capacitance 'C'. The switch 'S' is kept closed to point–1 for long time, then at timet = 0 switch'S' is shifted to position–2 from position–1 ColumnI Column II (A) Final charge on capacitor C1 (P) 2 CV 3 2 (B) Work done byBatteryhaving emf V2 = V (Q) 2 CV 3 4
- (C) Final charge on capacitor C2 (R) CV 3 2 (D)Amount ofheat produced inthe circuit after (S) 3 CV shifting the switch 'S' from (1) to (2) [Ans: (A) S; (B) P; (C) R; (D) P] [Sol: Initial charge on C1 = CV KVL C CV q1 + C q1 + C q1 = V q1 – CV + 2q1 = CV 3q1 = 2CV q1 = 3 CV 2 Final charge on C1 = |q1 – CV| = 3 CV 2 – CV = 3 CV work done by battery (V2) = q1 × V = 3 CV 2 × V = 3 CV 2 2 final charge on C2 = 3 CV 2 Ui = 2 1 CV2 Uf = 2 1 9 iV 4 2 × 2 + 2 1 9 CV2 = 18 1 18 8 CV2 = 2 CV2 H = 3 CV 2 2 ] Q.2 Match the following. (Here gas is ideal, P = pressure, V = volume, = ratio of molar specific heat at constant pressure&constant volume) ColumnI ColumnII (A)AdiabaticBulkmodulus ofgas (P) – V P (B)Slope ofP–V graph inisothermal (Q) 1 2 process (C) Degreeof freedom (R) P (D) Molar heat capacityat constant pressure (S) 1 divided by R [Sol: (A) PV = C dP d PV = 0 V dP dV V · P 1 = 0
- dPV PdV = 0 Bulk modulus = – V / dV dP = P (R) (B) PdV + VdP = 0 Slope = dV dP = V P (P) (C) = 1 + f 2 f = 1 2 (Q) (D) 1 R 1 R R CP (S) ] PART C Q.1 Asampleofmonoatomicidealgasoccupies5.00Latatmosphericpressure and temperature 300 K (pointA). It is heated at constant volume to 3.00 atm (point B)then it is allowedto expand isothermallyto 1.00 atm (point C) and at last compressed isobaricallyto its original state. If the gas is monoatomicthencalculatethetotal amountofheat(injoules)giventothe gas during whole cycle. (Here R = 8.30 Joule/mole0K; Patm = 105Pa; 1L = 10–3 m3; ln3 = 1.1) [Ans: 650 Joules] [Sol: A Isochoric B Isothermal C Isobaric A P1 = P0 = 105 Pa P2 = 3P0 P3 = P0 P4 = P0 V1 = V0= 5 x 10-3 m3 V2 = V0 V3 = 3V0 V4 = V0 T1 = T0 = 300 K T2 = 3T0 T3 = 3T0 T4 = T0 T1 = T4 = T0 U = 0 Q = W = areaABCA= (Area under BC curve) - (Area under CAcurve) = (nRT0) ln 2 3 V V - P0 (V3 - V V4) = 3P0V0ln 3 - 2 P0V0 = P0Vo (3ln 3 - 2) = 650J Q = 650J] Q.2 Inthecircuit shown in figureswitch 'S'is closedforlongtime, then at time t = 0 it is again opened. Find the charge on capacitor (in C) at time t = 6s (here e–1 = 0.37) [Ans: 7.4C] [Sol: At node X 2 0 X 4 30 X X – 30 + 2X = 0 3X = 30 X = 10 At nodeY 4 0 Y 2 30 Y = 0 2Y – 60 + Y 3Y = 60 Y = 20 Potential differenceacross capacitor
- = Y – X = 20 – 10 = 10 q0 = CV = 2 × 10 = 20 C When switch S is open charge on capacitor at t = t Req = 3W q = q0 q Re C t e q = 20 3 2 6 e = 0.37 × 20 = 7.4 q = 7.4 C ] Q.3 ArodoflengthL=1mwiththermallyinsulatedlateralsurfacehasuniformcross-section.Thecoefficient ofthermalconductivityofthematerialoftherod varieswithtemperatureas K T 1 .Theendsoftherod are maintained at temperature T1 = 200C & T2 = 800C. Find the temperature (in0C) at the mid point of rod. [Ans: 400C] [Sol: dx dT kA dt dQ = dx dT A T C dx CA dQ = T dT 1 0 0x C = 80 20 T ln C0 1 = –ln4 ..........(1) Similarly 2 / 1 0 0x C = T 20 T ln C0½ = –ln 20 T (2) Dividing2= 20 T ln 4 ln 2 20 T = 4 20 T = 2 T = 400C ] Q.4 Fiveidentical largeconductingplates each ofareaAareplaced parallel to each other at separation 'd'. Plates 1 & 3 are connected by thin conductingwire&plate2&5areconnectedbyanotherthinconducting wire.Thejunctionof plate1&3and plate-4areconnectedbyabattery having emf'V'(as showninfigure). The system is in steadystate. Now Plate-1ismovedupwardslowlysothatitcomeincontactwithPlate2. Calculatethe amount ofwork done (in J) bybatteryduringmotion of plate-1. (Here d A 0 = 20F, V = 10 volts) [Sol: C = d / A 0
- When 4 and5 are not connected the equivalent Ckt diagram Cleft = 3 C 5 Charge given by batteryq1 = Ceff · V = 3 CV 5 When 4 & 5 are connected, the equivalent ckt diagram Ceq = 2C q2 = 2CV work done by battery = V CV 3 5 CV 2 = 3 CV × V = 3 1 CV2 = 3 1 × 20 × 100 = 3 2000 J = 666.67 J ]

Anúncio