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CBSE Class 12 Pre Board 2021 : Physics - with ANSWERS / MARKING SCHEME (Kendriya Vidyalaya (KV), Kanpur Cantt)

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PRE-BOARD EXAMINATION-2020-2021 SUBJECT- PHYSICS Pre Board-1(2020-2021) Subject- Physics Marking scheme 1- (1/2) E= 10/9 V/ m (1/2) 2- 1:1:1 (1) 3- Rectangular shape because of constant rate of change of area OR Plate A will be positive (1) 4- Intensity will remain same Or 18 5- M = (1) (1) (1/2) (1/2) 6- Correct explanation (1) 7- Not correct with reason (1) 8- Vertical component of earth magnetic field= 25x 10-4T Induced emf E = 3.1V (1/2) 9- Velocity of photoelectron increases (1) 10- For electron -Z axis (1/2) For proton +Z axis (1/2) 11- a (1) 12- b (1) 13- a (1) 14- a (1) 15- (i) a (1) (ii) a (1) (iii) d (1) (iv) a (1) (v) a (1) 16- (i) b (1) (ii) c (1) (iii) c (1) (iv) b (1) (v) a (1) 17(1) W = -2x10-9x8x10-3x9x109x( (1/2) (1/2) (1) W= 12J OR (1/2) ( ) X=10cm 18- correct figure (1) (1) (1/2) (1/2) (1/2) r n2 (1/2) or (i) Yes, they are the isotopes of ther same element because they have the same atomic number (1/2) X is more stable than Y (1 ) (ii) 19- E= 9-5= 4V R= I=E/R = 0.5A (1) If current flowing through 3 resistance be I1 then 3I1=6x(0.5-I1) I1=0.33A (1) 20- Diffusion and Drift Correct diagram (1) (1) Or D1 is forward biased and D2 is reverse biased Equivalent resistance R= 2+1= 3 (1) I= V/R= 6/3=2A (1) 21- Electric field intensity correct definition, unit and vector quantity Electric flux correct definition, unit and vector quantity 22(1) (i) Fringe width increases (1/2) (1) (1) (ii) Fringe width increases (1/2) 23- (i) by radioactive decay (iii) 24- Currect justification 25- (i) electrostatic shielding (ii) ciorrect diagram 26 =h/p mv= h/ hr/ = (1) (1) (1+1) (1) (1) (1) (1) 2 r=n (1) Or E1= -13.6 eV E2= -3.4eV E3= -1.51eV E4= -0.85eV E3-E1=12.09eV E4-E1= 12.75Ev Upto E4-E1 energy level the H atoms would be excited. Lyman series 1 = 1.215x10-7m Balmer series 1 = 6.56x10-7m (1) (1) 27- (i) (a) = =50 rad/s (1) (b) Z = = 40 Irms= 5A I0=Irms = 7.07A (1) (ii) correct explanation (1) Or (i) E = (100sin314t)V I=I0sin(314t+90)=I0cos314t I0=E0/XC = E0 c I= 20cos314t (2) (1) Angular frequency of power p=628 rads-1 fp= 100Hz (1/2) 2 -6 (iii) U0= CE0 /2= 637XX10 X1002/2 = 3.185J (1/2) 28- Correct diagram (1) M=-f0/fe (1) The diameter of objective is kept large to increase the intensity of image and resolving power of image. (1) 29- (i) forward and reverse characteristic (1 ) (ii)(a) r=20 (b) r= 4x106 (1 ) (ii) 30- (i) I = = 2A All cells are connected inseries (1) (ii) 4 cells are connected wrongly by Shikha (iii) R=0 IMAX= E/r= 20/0.5= 40A (1/2) 31- any two differences (2) M= L= 7.50cm (1) (1) U0= -1.5cm (1) (i) (ii) Or S1P= S2P= S2P-S1P = (2n-1) /2 (n=1) (2) Correct solution and D= 0.404 Correct reason for coherent source (1) 32- correct solution Or Correct definitions (1 ) (2) (5) (1 ) Corrct diagram (1 ) Correct relations (2) 33- different ways to produce emf (2) Correct derivation (2) Lenz s law and Fleming left hand rule (1) Or (i) Correct derivation of instantaneous current (ii) Correct derivation of average power (iii) Phasor diagram (1) ------------x------------x---------------- (2) (2)

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