8.PRACTICE TEST-1 (Solution-Paper-2)

PAPER - 2 PART-I (PHYSICS) 1. Pressure at 'A' from both side must balance. Figure is self–explanatory.  h g = h g 2 1  sin (45º + ) = R [cos – sin]  [cos + sin] = [cos – sin]    tan =    2  20 sin 37 2. T = 10 12 3 = 4 × 5 12 = 5 sec. Range = R = 5 × (20 cos 37° + 10) 12 4 5 × (20 × 5 12 + 10) = 26 × 5 = 62.4 m 3. If A moves down the incline by 1 metre, B shall move up by of A will be 2v. From conservation of energy: Gain in K.E. = loss in P.E. 1 metre. If the speed of B is v then the speed 2 1 (2v)2 + 2 A 1 v2 = m g × 3 2 B A 5 1 – mB g × 2 Solving we get v = 4. As when they collide vt  Ans. 1  72v 2  2 2  25R   R = vt  t = 5R 6v   Now angle covered by A =   vt R 11 Put t  angle covered by A = 6 5. Force on table due to collision of balls : dp Fdynamic = dt = 2 × 20 × 20 × 10–3 × 5 × 0.5 = 2 N Net force on one leg = 6. Case Ι 1 (2 + 0.2 × 10) = 1 N 4 R ×  g 5 S = 4 =     × 4 5    Rg = 16  Case II 4  2 16 1 = 6  1 = /13 ( – 1) 2 so decrease in current to previous current  / 5   /13 =  / 5 8 = 13 Ans. 7. E = 2 Q 2A   2A   E = 3 Q 2A  3 Q 3 Q E = 2 Cd  Ed = 2C = V (ii) F = EQ/2  2Q  (Q) Q 2 F =  2A   × 1 Q 2  A   F = A   1 1  3Q 2 9 Q2 (iii) Energy =  E2 Ad = 0   A d = 2 0 2  2cd 8 C 8. When capcitance is removed L tan  = R or L = 100 tan 60º (1) when inductance is removed 1 tan  = (C)(R) 1 or C = 100 tan 60º (2) so z = R = 100   = v/R = 200/100 = 2A Power P = 2 R = 4 × 100 = 400 W h  9. The relation between the stopping potential and the frequency is h = eVS +  or VS= e  – e b From the graph slope = a b h  a = e  h = ab e From the graph y intercept = b   b = e   = b e 10. Let the normal force between the block and the ball be N. a N ma(pseudo force) mg For the block, from Newton’s IInd law , we have N = Ma = 2ma For ball (with respect to the block), from Newton’s IInd law , we have N + ma = Solve the two equations. mv2 R 11. The stress is max at the uppermost point and is equal to the weight of the rod divided by the area.At the highest ALg point stress = breaking stress =  = weight of rod/area of cross section of the rod = A = Lg   L = g . The stress decreases linearly to zero at the lowest end. 12. Before filling the dielectric medium the equivalent capacitance of the circuit is Ci = 8/3 F . After filling the dielectric medium the equivalent capacitance of the circuit is Cf = 4 F q = C V = (4 – 8/3) 12 = 16 C. Energy taken from the cell = charge flown through the cell x cell voltage= 16 x 12 = 192 J 1  1  8 Ui = CV2 =  2  x 3 x 122 =192 J 2   Uf =  1  CV2 =  2  x 4 x 122 = 288 J. 2   Therefore increase in electrostatic energy stored in the capacitors = 96 J. 13. (NBA)i = c C (6  105 )    i = NBA =   4 2 10 1   10   i = 30 mA 30 So current corresponding 1 part = 10 = 3 mA. 14. From momentum conservation mu = mvcos 300 + mvcos 300] Initial KE = 1/2 mu2 mu2 Final KE = 6 Loss in KE = mu2 /6 Therefore mu2 /6 = 10.2 eV 1/2 mu2 = 30.6 eV 15. n2  n1 = n2  n1 v v 1  1.5 v  2R R 1 1.5 =  R   = –4R m = n1v n2u 1.5  (4R) = 1 (2R) m = 3. 16. g – 2 = 1 × a 12 – 5 = g = 4 × a a = 1 m/s2  = 0.3 Ans. 10 = 3 Ans. 17. T = mg TR = Icos MgR a = MR2 = 2g 2 R 2  = 2 2R = 4g 2R2 L = 4g  / 2 18. p =  2dp sin  0  / 2 =  2 (0 sin  Rd . 2R) sin  0 [direction – ˆj ] x  / 2 = 20 R2  (1 cos 2)    d = 2 R2   = R2   0 19. (A) Speed of charged particle cannot be changed by magnetic force because magnetic force does no work on charged particle. Only electric field in case (p) and induced electric field in case (r) can change speed of charged particle. (B) Magnetic field cannot exert a force on charged particle at rest. Only electric field in case (p) and induced electric field in case (r) can exert force on charge initially at rest. In case (r) after the charge starts moving even the magnetic field can exert force on charge. (C) A charged particle can move in a circle within uniform speed due to uniform and constant magnetic field in case . Even within a region of non uniform magnitic field, at all point on a circle field may be uniform for example on any circle coaxial with a current carring ring (D) A moving charged particle is accelerated by electric field and also accelerated by magnetic field (provided v is not parallel to B). u2 sin2  100 3  20. (A) R = g 2(10) 5 3m 1 (B) 11.25 = – 10sin 60º t + (10) t2  5t2 – 5 2 t – 11.25 = 0 5 3  t = 25(3)  4(5)(11.25) 10 = 5 3  10 3(10) = 15 10 3  3 3 2  3 (C) R = 10 cos 60     = 7.5 m  t = 2usin 30º  1  2(10)      2 sec. 30º 30º g cos 30º g gcos30º  3  10      1 10 3  2  1  1  4 10 20 R = 10 cos 30º t  g sin 30º t2 =    (10)  2 3 2 2 3 = 10   m (D) T = 2 2(10)  2(10)      3 3 sec. gcos 30     1 R = 2 g sin 30º t 2  1  = (10)  2  16 40 3 = 3 m 2  