Physics-4.UNIT - 06 TEST SOLUTOIN

UNIT - 06 TEST SOLUTOIN 1. (A) The direction of (about the center) is perpendicular to the plane containing the circular path. Both magnitude and direction of the angular momentum of the particle moving in a circular path about its center O is constant. 2. (B) The linear momentum (L) is conserved, since τext is zero. Let Ii = I0 ∴ If = 2I0 ωi = ω0 ∴ ωf = ω(say) ∴ I0ω0 = 2I0ω ⇒ = 3. (C) Here VC = VC Vp = VQ = 4. (D) x = cos ; y = sin = 1 x2 + y2 = = Circle 5. (A) Iw = constant When ballet dancer then I will increase will I = 6. (D) mgR = –mgR + 2mgR = w = 7. (D) mg sinα – T = mA T R = Iα, A = R T = Mg sinα– =Ma So a=1.4m/s2 8. (D) dm = dx dy = Consider the y coordinate of c.m. of strip = = = = = = The system is symmetrical in x and y So, x = 9. (C) 10. (A) by conservator angular momentum about joint anti clock wise 11. (C) & V0 = r0 w0 12. (A) N = mLα 13. (A) Accl of 0 (i.e. centre of mass) a = 14. (B) = 0.96 kg m2 15. (A) Rotational kinetic energy Joule Joule = Joule 16. (C) IB=IA + Md2 17. (D) 18. (A) 19. (C) 20. (C) 21. (B) 22. (C) For conservation of angular momentum 23. (D) 24. (D) fr = Ma ….(i) fr R = Iα = …(ii) fr = μ mg …(iii) By (i) and (iii) A = μg μ mg = ⇒ will none of the option A, B or C 25. (A) 26. (C) When body will rolls ultimately in forward direction then at that time if its velocity is v0 and angular velocity is ω then V = Rω ….(i) By conservation of angular momentum about S. ⇒ we know that finally. ⇒ ⇒ MV0R – Iω0 ≥ 0 ⇒ V0 ≥ = = So V0 min = 27. (B) And by above = mVR + ⇒ ⇒ At the time of rolling ⇒ V = Rω0 By conservation of angular momentum about S during collision. Lsi = Lsf = I S ω1 = ω1 = After collision by conservation of energy for ⇒ V0 = 2Rω0 = 28. (C) If S is perfectly rough then N and fr both will an impact force and after collision velocity of point of contact of disc with S will become 0 and disc will roll purely about S. If S is perfectly smooth then No fr will act and no change in ω0 will takes place. By conservation of energy after collision and when it will on the upper plane. sinθ = 3/4 v = R Now at upper plane By conservation of angular momentum about k Lki = mV1R + Lkf = mV2R + = mV2R + = ⇒ V2 = 29. (D) v0cos2θ = 2v0 sinθ cosθ = v0 sin2θ = V3 > V1 So finally sphere will perform projectile with velocity = and velocity ↑ = and when it will again collide upper surface then its nel. will be = ↓ and sin θ = 3/4 cosθ = Conservation angular momentum during collision. LSi = LSf = mVR + Iω LSi = LSf So Ans. (D) 30. (B) If body go t in particle direction only When eVcosθ × cosθ = vsinθ sinθ ⇒ e = tan2θ = So, its not possible

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