CLASS XII-FIRST COMMON PREBOARD EXAMINATION

PHYSICS - Scoring key

1 No. Electric potential may be zero or constant at that point 1 mark2. Vrms =V0/2 ½ mark V=220 2 sin 100t

½ mark

3 Towards left. 1 mark4 gamma rays, X- rays, micro waves, radio waves 1 mark 5 No change 1 mark6 5V 1 mark 7 σ decreases 1 mark 8 AND Gate & its truth table 1 mark9 R’ =R1R2/R1+R2=2Ώ

Net emf E=E2-E1=4 V I=0.5 A 2 marks

10 x=2/3 m or -2 m x=-2 not valid so intensity is zero at 2/3 m from Q1 2 marks

11. If same p.d is applied, current = I1 and I2 will flow through the wire, when its temperature is kept at T1 and T2 resp. 1 mark Resistance is more for T2 1 mark12. Tan = Bv/BH

=600 1 mark BH=Bcos = 0.2 Tesla

1 markOr

Any two points 2 marks13. a) E0=nAB = 720 V 1 mark I0= E0/R=1.44 A

b) P= E0I0 = 1036.8 W 1 mark

14. Proof of I= I0 sin (t-/2) 2 marks15. The electric and magnetic field vectors oscillat perpendicular to each other and to the direction of propagation of the wave.

1 mark Diagram 1 mark16. Labelled ray diagram 1 mark

M=f0/f e

1 mark

17.i) Decreases ii) Increases 2 marks18. Point to Point communication -The communication over a link between a single

transmitter and receiver e.g: Telephone ½ + ½ mark Broadcast- Large number of receivers is linked to a single transmitter. E.g. Radio.

1 mark19. (i) By increasing resistance R, the current in main circuit decreases, so potential gradient decreases. Hence a greater length of wire would be needed for balancing the same potential difference so the null point would shift towards right(i.e towards B)

1 mark (ii) By decreasing resistance S, the terminal potential difference V=E/(1+r/s) is obtained at smaller length. So the null point would shift towards left(i.e towards A)

1 mark20. Self inductance of a solenoid of length L=0n2AL, where n is no. of turns per metre.

L1=0n12A1L

L2=0n22A2L 1 mark

If I, is the current in outer solenoid, then magnetic field at axes B1=0n1I1

Magnetic flux linked with the secondary coil 2=(n2L) B1A2=(n2L)(0n1I1)A2=0n1 n2 LA=I1

M= 2/I1 = 0n1 n2 LA2

L1L2= I1(0n12A1L)(0n2

2A2L) L1L2=0n1 n2 L A1 A2

M/ L1L2= A2/A1, A2<A1, A2< A1 A2

M< L1L2

21. Statement 1 markProof with neat diagram. 2 marks

Or Statement 1 mark

Brewster’s law. Proof of r+ip=900 2 marks

22. Resolving power of a microscope= 2sin

a) Resolving power increases 1 markb) resolving power decreases. 1 markc) no change . 1 mark

23. Neat labeled diagram 1 mark Explanation 2 marks

24. Figure 13.1-Page No.444 1 mark Conclusions regarding fission and fusion 2 marks

25. En=-13.6/n2 eV E1=-13.6eV(for n=1) E4=-13.6/42 = -13.6/16(for n=4) Energy of photon = h=-0.85 – (-13.6)=12.75 eV 2 marks hc/=12.75 eV =974 Å

=c/ = 3 × 108 / .974 × 10 -7 =3.1 × 10 15 Hz 1 mark

26. i) Figures (a) and (b) 14.28-Page 492 2 marks ii) The base is made thin so that most of the holes find themselves near reverse-

biased base-collector junction and so cross the junction instead of moving to the , base terminal. 1 mark

27. (a) modulation = 10/20 =Em/Ec=0.5 1 mark (b) The side bands are at (1000+10 KhZ) 1 mark =1010 Khz and (1000-10Khz)=990 Khz. 1 mark

28. (a) Explanation with diagram 1 mark

(b) Proof of C= E0A d-t+t/k 4 marks

OR(a) Statement of Gauss’s theorem 1 mark(b) Derivations of (i) field outside the shell 2 marks E = q r 4E0r2

© Field inside the shell- Proof-E=0 2 marks

29. Principle 1 mark Construction 1 mark Working 2 marks Importance of radial field 1 mark

OR Principle 1 mark Construction 1 mark Working 2 marks Not suitable for accelerating electron-reason 1 mark

30. (a) Lens formula with neat ray diagram 2 marks(b) Three ray diagrams 3 marks

ORAny two conditions with neat diagram 1 markDerivation of expression for 4 marks