8.Subject Test 1 - 2 (Solution)

SOLUTIONS TO SUBJECT TEST - 1 (ST-1) PAER -1 1. V = V1 + V2 + V3 1 Q 1   2Q  1  3Q   1  2Q   = 4 . R + 4  R  + 4  R  = 4 .  R  0 0   0   0   2. VB – VA = – Ex dx = – [Area under Ex – x curve] VB – 10 = – 1 .2. (–20) = 20 2 VB = 30 V. 3. As soon as the field changes, current is induced in the anticlockwise direction. Now direction of M and B are parallel thats why torque on coil is zero. 4. The magnetic flux must remain constant   = B0ab = B0 1 kt bx where x is as shown  x = a(1 + kt) or v = dx = ak dt 6. Initially the potential at centre of sphere is VC = 1 Q 4 0 x + 1 2Q 4 0 x 1 3Q = 4 0 x After the sphere is grounded, potential at centre becomes zero. Let the net charge on sphere finally be q.  1 q  1 3Q = 0 or q = – 3Q r 4 0 r 4 0 x x  The charge flowing out of sphere is 3Qr . x 1 d 7. i.(t) = – R dt = 2bt – 3at2 di i is maximum when dt = 0 2b or t = 6a 2  6 = 6  2 = 1 at t = 1 s, i is maximum its value = 12 – 6 = 6A 3KQ2 8. U1 = a 3kQ2 where a = side of equilateral triangle Uf = a + Q where q is the charge brought at the center U = 0  q = . 9. Net potential of the sphere due to the induced charge is zero. Therefore potential is due to the point charge only, equal to Q potential at the centre of the sphere. + KQ Vsp = d 1   2 10. Just before opening the switch, the current in the inductor is /R. Energy stored in it = L  . 2 R This energy will dissipate in the resistors R and R in the ratio   1 & 1 . 11. Charge on a1 Charge on a2 = (r1 = (r2 )  )  r1 1 2 R1 R2 Ratio of charges = r2 K [ (r1  )] K E1, Field produced by a1 = E2, Field produced by a2 = as r2 > r1 Therefore E1 > E2 2 1 K r2 = r1 i.e. Net field at A is towards a2. K.(r1) V1 = V2 = r1 K.(r2 ) r2 1 = K  = K   V = V2 . 12. (A) E = 2 CV2 As potential difference source between the plates is connected, p.d. remains constant. But capacitance C becomes KC hence energy stored is increased by factor K. V (B) Electric field d is not changed. (C) Charge on each plate is increased by factor K hence force between them increases by factor K2. For effect of the medium, they must completely lie in the medium. (D) Q = CV Hence charge becomes KQ as C becomes KC and V remainin unchanged. 13. The potential at surface, 5 cm from surface and 10 cm from surface outwards is V = KQ s R .... (1) KQ R  5 = 100 .... (2) KQ R  10 = 75 .... (3) From Equation 2 and 3  R = 10 cm  From equation 2 100  15  102 Q = 9 109 5 = 3 × 10–9 C  B is false V = KQ surface R 100(R 5) = R 100 15 = 10 = 150 V  A is true V = 3 KQ = 225 volts  D is true centre 2 R KQ Esurface = R2 150 = 10  102 = 1500 V/m.  C is true 14. Rate of work done by external agent is : dw BL.dx dt = dt = BLv & thermal power dissipated in the resistor = e = (BvL)  clearly both are equal If applied external force is doubled, the rod will experience a net force and hence acceleration. As a result velocity increases Since ;  = e R On doubling ‘R’, current and hence required power becomes half. Since power is halved hence velocity is also halved as P = BLv 15. Arc AB =  r 3  mV = 3qB  T     T m Time 't' =  =  2 . 3   6 = 3qB .     16. Potentials are indicated in figure 2 10 5 5 30 Current in 2 = 10  (–5) 15 2 = 2 = 7.5 A, leftwards 5 • -15 10 0 0 • 5 5 -5 15 5 Current in 30  = i1 10 (15) 30 25 = 30 5 = 6 A, downwards assume 5 5 i2 = 9 17. Just before S2 is closed the current in the inductor is /R. It will not change just after S2 is closed due to property of inductor. At the moment S2 is closed  Let X be the potential of the junction then i1 X /R i2   i1 + i2 = R L   X  3  X   R 2R R  X = –  Now using Kirchoff's law di 3 + L – 2 = 0  di    = – 10 A/s dt dt L A 18. Number of photons striking per second = N = h Area A here is the area perpendicular to the direction of intensity or direction of energy flow. Consider a ring of radius x and width dx on the disc. Intensity  on the ring due to source is  = P 4r 2 now, dA = dA cos  dA = 2xdx Gives, dA = (2xdx) cos  A photon will exert force F as shown, only the Fcos  component will remain and F sin  will cancel out P Nh  dA   h as we integrate on the ring. t =  =  h  as only Fcos component of force remains  dA   h dF =   h F  × cos    P  (2dx cos ) h  dF    4(4R2  x2 )  × hC ×  cos   r =  0 0  P        solving we get F = 20c 19. Let i1 and i2 be the current through L and R at any time t  i = i + i2  B𝑙v R = i2 and B 𝑙 v = L dv di1 dt i  B𝑙v  Force on conducting rod = m d t =  i 𝑙 B =   1 R  𝑙 B  m d v =  𝑙 B i dt  B2 𝑙2 R v d t  m  d v =  𝑙 B  i1  B2 𝑙2 d t R B2 𝑙2  v d t m (vf  u) =  𝑙 B Q – x R (vf = velocity, when it has moved a distance ' x ' )  Q =  B2 𝑙2 x m (v B𝑙  u) = 1C. 20. Let nth minima of 400 nm coincides with m th minima of 560 nm, then  400   560  2n  1 7 14 (2n – 1)  2  = (2m – 1)  2  or 2m = 5 = 10 =.........      1 i.e. 4th minima of 400 nm coincides with 3rd minima of 560 nm. (2  4  1)(1000)(400 106 ) Location of this minima is, Y1 = 2  0.1 = 14 mm Next 11th minima of 400 nm will coincide with 8th minima of 560 nm. Location of this minima is, (2  11 1)(1000)(400  106 ) + Y2 = 2  0.1 = 42 mm  Required distance = Y – Y1 = 28 mm = 7X Hence X = 4 SOLUTIONS TO SUBJECT TEST - 1 (ST-1) PAER -2 1. Charge on capacitor before insertion of dielectric slab = 100 µC Charge on capacitor after insertion of dielectric slab = 300 µC Increase in charge on the capacitor = 300 – 100 = 200 µC charge increases very slowly that's why heat will not be produced Energy supplied by the cell = increase in stored potential energy + work done on the person who is filling the dielectric slab.  Heat produced is zero so (D) option is also correct. 25 V 25 V 25 V 25 V 2. Power supplied by 20 V cell = (–1) (20) = – 20 W (as the cell is not supplying the power, it is eating the power (getting charged) 3.  =  dt      1  K.E. = 1 2  1 2  K1 K2 = K1 mr 2 2 MR 2 2 m  r 2 K =   2 M  R  K1 + K2 = U m  r 2 K2    K2 = U M  R  UMR2 K2 = K1 = mr 2  MR2 Umr 2 mr 2  MR2 U =  r 2 2 0 2R  2r 2 = 4R 4.  – 0.5  + 6 – 5 2 = 0 –   q  3  2 2 2  = 2A = 0 q = 2C. 5. tani = 1.6  4 1.2 3 angle of incidence i = 53° 4  = 3 = sin 53 sinr sinr = 4  3 = 0.6 5 4 r = 37° tanr = 3 x' 4 = h  x' = 3m x = (1.6m + 3m) = 4.6 m 6.   iR1  R 2   2 x (2  4)  12  ACD  6V ; DEA  6V The equivalent circuit is as shown VA + 6 - 2(2) = VD VA + 2 = VD  VD - VA = 2 volt 9. Consider two small elements of ring having charges +dq and – dq sym- metrically located about y-axis. The potential due to this pair at any point on y-axis is zero. The sum of potential due to all such possible pairs is zero at all points on y-axis. Hence potential at P(0, R ) is zero. 2 x´ x y´ 10. Consider two small elements of ring having charge +dq and –dq as shown in figure. The pair constitutes a dipole of dipole moment. dp = dq 2R = ( Rd) 2R The net dipole moment of system is vector sum of dipole moments of all such pairs of elementary charges. By symmetry the resultant dipole moment is along negative x-direction. x´ x  net dipole moment = – = – 4R2  ˆi  / 2 (dp cos ) ˆi    / 2  / 2 (2R2 cos d) ˆi  / 2 y´ 11. From the passage: b  cot /2 therefore cot; /2= 0.  /2 = /2   = . 12. + emission: x  y + + +  n  P + + +  Q1 = (Mx – My – 2me)C2 Electron capture: x  y +  + x–rays p + e–  n +  Q2 = (Mx – My)C2 Q1 > 0 implies Q2 > 0 but Q2 > 0 does not necessarily mean Q1 > 0. 13,14 Let potential of point P be zero  Just after switch closed V = +3 volts. Also VA = +1 volt  Current through R VQ  VA just after switch K 3  1 is closed is R1 = 2 = 1 ampere Any time after switch K is closed VA = 1 volt, VB = 3 volt  V – VA remains constant. After the system has finally achived steady state, potential difference across capacitor. R1 R2  E1 E2  5 VP – VQ = R  R    = volts 3 1 2  1 2   Charge on capacitor = C × 5 3 = 50 C 15. Let at an instant vR = (VR)m cost  2 = 4 cost cost = ½  cost = 60°. Since VL is 90° ahead of VR vL = (VL)m cos (t + 90 ) = – 3 sint = – 3 sin 60° = – 3 cos 30°  | (V ) | = 3 cos 30° 16. From phasor diagram (VS)m = tan  = VL )m  3   = 37° = 5 volt. VR )m 4  | v | = | (V ) cos (t + 37°)| S S m = 5 | cos (60° + 37°)| = 5 | cos 97°| = 5 cos 83° 27 (A) Activity of the sample II becomes half in minimum time. Hence it has maximum disintegration constant. (B) Activity of the sample III takes maximum life to become half therefore it has maximum half - life. (C) Parent nuclei will be left maximum in the sample, for which half life is maximum i.e. minimum decay. (D) It can not be compared without information about atomic weight as energy radiated will depend upon no. of atoms, not upon amount of substance. 28. (A) Electrostatic potential energy = 1 4 0 (Q)2 2a Q2 = 8 0 a 1 (Q) (Q)  (Q)2  3 Q2 (B) Electrostatic potential energy = 4   5a / 2 2(5a / 2)  = 20   a 0   0 1 3Q2 3 Q2 (C) Electrostatic potential energy = 4 0 5a = 20  0 a 1  3Q2  (Q)2  (Q) (Q)  27Q2 (D) Electrostatic potential energy = 4   5a 2(2a) 2a  = 80  a 1 1 29. E  n2 , V  n and r  n2 0   0 1 1 1 (A) Epr   n2 n × n2 or Epr  n (B) (B) p  1  n2 or p  n E n E 1 (C) Er  n2 (D) Pr  1 n × n2 or Er is independent of n × n2 or pr  n 1/ C 30. (a) tan  = R   =  4 , current leads source voltage because reactance is capacitive (b) Pure inductive circuit  = /2 , current lags behind source voltage because reactance is inductive (c) as R = 0, tan    ,   /2 , current leads source voltage because reactance is capacitive (d) tan  = L = 1   = R  , current lags behind source voltage because reactance is inductive 4