4.Solution Combine Test-2 (Paper-2)

SOLUTIONS TO COMBINED TEST - 2 (PAPER - 2) h 1.*  = p p is the same p = mv vp  me ve mp p2 p = mK = 2 2. The image of a point closer to the focus will be farther. As the transverse magnification of B will be more than A, the image of AB will be inclined to the optical axis. 3. Induced emf  1 2  e = vB𝑙 = ( 3 t) . B  2 ( 3 )t (2 tan 30) = B t3 e1sec e10 sec 1 = (10)3   3 Bt4 4 e2sec = 24 Volt → → 4. Using – e (V B) for the region outside the plates, direction of magnetic field can be found. Inside the plates, net force on the electron is zero hence electric force is opposite to that of magnetic force. Direction of electric field between the plates is opposite to that of direction of force on the negative (electron) charge. 5.  net  qm 0 = 0 as qm = 0 → →  E.ds = 0, but we cannot proceed further, as E cannot be taken out of the interval. Hence we cannot say E = 0. Infact electric field due to a dipole is non–zero at any point at a finite distance from the dipole. 6. 2MV0 = 2MV1 + MV2 1 = e = 2 (V2 cos   V1 cos) V0 cos V1 = (2 – e) V2 = V0 3 2 V = (1 + e) 2V0 = V 2 3 0 1 K.E. = K.E. = 2m V 2 2 1 V 2 2m × + 2 4 1 1 × M × V 2 2 Loss in K.E. = mV 2 . 4 7. If the angular displacement is  for 0.5 seconds, then the magnitude of change in velocity is  V =  V = 2 × 4 sin  or 2 = 2v sin 2  = 2  &  =  6 3   =  0.5 = 2 3 rad/s v R =  = 4  3 2 6 =  m 2 8 and centripetal acceleration = v ×  = 4 × 3 8. At resonance XL = XC and Z = Zmin = R = 3 m/s2 . XL = L and XC = C 9 & 10. If f is decreased then ‘’ will decrease and hence XC will increase therefore at f < fr, circuit behave as capacitance. Also, VL and VC always difference in phase by 180°. When it is connected to AC source 2.4 =  L = 0.08 H When inductor is generted to DC source  = V ; 4 = 12 11 & 12. R R  R = 3  From graph it is clearrest current is zero at l = 20 cm hence balance point is at 20 cm Now exp of the call = 6  20 100 = 1.2 V 13. P = V is where V = accelerating voltage is = saturation photocurrent is = = = 0.01 A  Power = 100 watt 14. Maximum energy of incoming electron = –  + eV = eV = 10,090 eV ; min = = 1.23 Å p 15. r = qB > d  p > qBd = 2eBd for  particle p 16. r = qB < d  p < qBd = eBd for proton/electron (neutrons pass over to region  undeflected) 17 The initial charge on capacitor = CVi = 1 × 2 C = 2 C The final charge on capacitor = CVf = 1 × 4 C = 4 C  Net charge crossing the cell of emf 4V is qf – qi = 4 – 2 = 2 C The magnitude of work done by cell of emf 4V is W = (qf – qi) 4 = 8 J The gain in potential energy of capacitor is U = 1 C(V 2  V 2 ) = 1 × [42 – 22] J = 6 J 2 f i 2 Net heat produced in circuit is H = W – U = 8 – 6 = 2 J J 18 Drift speed Vd = ne i = neA i = V R L where R = A V E = L and P = I2 R 19. For all types of wave, sound wave, light wave, string wave the term related is frequency. Which is given only in one option. Other phenomenon are properly matching. Photoelectric effect proves photon character of light.  rays can only be produced from nucleus. In case of k capture x rays are emitted. 20. P.E. = – GMm  K.E. = r 1 mV2 2 Total energy =  GMm r + 1 mV2 2 T.E. = 0 if  GMm  1 mV 2  0  v = r 2 For v < for v > , T.E. is – ve , T.E. is + ve