3.Solution Combine Test-2 (Paper-1)

SOLUTIONS TO COMBINED TEST - 2 (PAPER - 1) 1. Capacitor behaves as open circuit (when charged completely) after a long time. 2.  is RC i.e. time constant. After one time constant, capacitor looses 0.63 part of its initial charge.  t   RC  1 qfinal = q0 e   = q0 e |q| = q – q e–1 = q 1 1  = 0.63 q . 0 0 0  e  0 3. Even for non uniform charge distribution field may be zero. E.g. when charges on diametrically opposite points are equal. 4. VL = 8V, VR = 6V, V = Power factor = cos  = VR = V = 10 V 6 10 = 0.6 5. Emf between a point and center is given by 1 BR2 . 2 Therefore emf between two points will be 1 B(R 2  R 2 ) where R and R are the radial distances of the 2 1 2 1 2 two points from the center.Since point C is farthest from the center and the point E nearest, the emf will be maximum between these points. 6. BE of X = 6A BE of Y = 6A – 2 + 1 = 6A – 1 [Because absorption of energy decreases BE and release of energy increases BE] In Y nuclues there are A + 1 nucleons. BE  BE per neuclon of Y nucleus = nucleon 6A  1 = A  1 1 7. XC = C and XL = L At resonant frequency  ), 1 ω0C = ω0L 1 At  <   ωC > L  XC > XL  The circuit is capacitive 8. Resultant voltage = 200 volt Since V1 and V3 are 180° out of phase, the resultant voltage is equal to V2  V2 = 200 volt q0 9. max current = = 30 mA 10. 11. 0 1 = 4  (Z1  1)2 (Z  1)2 = 4  Z = 2Z – 1 0 1 2 4 (Z2  1)2 1  (Z  1)2 = 4  Z = Z  1 . 2 2 12. (A) It is true when i = 0, then r = 0   = i – r = 0 (C) By Cauchy's formula n = a + light) b , medium has lowest 'n' for red colour (out of all the colours in white 2 V = C/n  V is maximum for red. (D)  = | i – r | 1 × sin i = n sin r for violet, n is maximum  r is minmum   is maximum 13. Consider a ring of radius x and thickness dx.  Equivalent current in this ring = 2  × charge on ring = 2 × (2  x dx) Q R2    2xQ  R   Q dB (due to this ring) = 0  dx  B =  0 dx. 2x  2 R2  0 2 R2 = 0Q . R = 2R 0Q 2R By right hand thumb rule magnetic field is outward. 14. A and C are standard results. Suppose charge flown through the battery is Q, then charge distribution will be as : A B The electric field in the region between A and B is = Q  2C 2A 0 – C  Q 2A 0 = 2Q  3C 2A 0  P.D. between the plates, 2Q  3C .d =   2Q  3C 1 =   2Q = 5 C  Q =  5C 2 2A 0 2 C 52C work done by battery = Q = 2 15. V = E – ir V = 10 – 5i  r = 5, E = 10V i = E  10 = 2 amp max r 5   13.6  1  1  16. Energy equivalent to line of lyman series  12 42  13.6  1  1      10.2 So according to the question  12 42  –     2.55 or max = 2.55. So the next integer is 3. 17. We know that the electric field due to induced charges at an exterior point like B is cancelled by that due to dq the charge kept inside the cavity. Therefore k r 2 1x10 6 = k 12 dq . or r 2 = 1 × 10–6 . 18. We have force per unit length , f  0i1i2 2r 0i1i2  22 = 0.1  i i = 106 We know that for a given product of two quantities their sum is a minimum when they are equal. So sum of currents is min. when i1 = i2 = 103 Now the magnetic field at a point 1m from both the wires is 19. Focal length of water–lens : 0i1 + 21 0i2 21 = 0103 21 + 0103 21 = 4 x 10–4 T. 𝑙  m 1 2 𝑙   1  = + 30 cm The combination of the lens and the mirror will act like a mirror, whose equivelent focus distance will be 1 feq = 1 = 1  2 fm f𝑙 1  2 feq  5  30 feq =  15 cm 4 Image distance is given by, 1  1 = 1 V U 1  1 feq 1 = . V (15)  15 / 4 V = – 3 cm. 20. For L–C1 system  = 6  100 3 and T1 = 2 = 100 sec. For L–C system  = 2 = 100 and T2 = 2 = 100 sec. Circuit L–C1 is on for 1.5  100 = T1 sec. 4  1 C V 2  the charge on capacitor C1 will be zero, so the entire energy   1 1  will go in inductor.  0.5  T2  1 C V 2  Now after 100 capacitor C2. sec. = 4 sec., 'i' in the inductor will be zero, so now all the energy   1 1  will go in the   1 C V 2 = 1 C V 2  1 (900 ) (1)2 = 1 (100 ) v 2 2 1 1 2 2 2 2 2 2  v = 3 volt.