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

Comments

Post a Comment

Popular posts from this blog

Physics-30.24-Physics-Solids and Semiconductors

UNIT 24 - SOLIDS AND SEMICONDUCTORS 1. SOLID STATE ELECTRONICS (SEMICONDUCTORS) (A) Energy bands in solids: (i) In solids, the group of closely lying energy levels is known as energy band. (ii) In solids the energy bands are analogous to energy levels in an atom. (iii) In solids the atoms are arranged very close to each other. In these atoms there are discrete energy levels of electrons. For the formation of crystal these atoms come close together, then due to nucleus-nucleus, electron-electron and electron-nucleus interactions the discrete energy levels of atom distort and consequently each energy level spits into a large number of closely lying energy levels. (iv) The number of split energy levels is proportional to the number of atoms interacting with each other. If two atoms interact then each energy level splits into two out of which one will be somewhat above and another will be somewhat below the main energy level. In solids the number of atoms is very large ( 1023). Hence eac
Read more

Physics-31.Rotational Mechanics

5.1 DEFINITION OF CENTRE OF MASS Centre of mass: Every physical system has associated with it a certain point whose motion characterizes the motion of the whole system. When the system moves under some external forces than this point moves as if the entire mass of the system is concentrated at this point and also the external force is applied at his point of translational motion. This point is called the centre of mass of the system. Centre of mass of system of n point masses or n particles is that point about which moment of mass of the system is zero, it means that if about a particular origin the moment of mass of system of n point masses is zero, then that particular origin is the centre of mass of the system. The concept of centre of mass is a pure mathematical concept. If there are n particles having mass m1, m2 ….m n and are placed in space (x1, y1, z1), (x2, y2, z2) ……….(x n, y n, z n) then centre of mass of system is defined as (X, Y, Z) where = Y = and Z = where
Read more

Planning to start your own coaching institute for JEE, NEET, CBSE or Foundation? Here's why not to waste time in developing study material from scratch. Instead use the readymade study material to stay focused on quality of teaching as quality of teaching is the primary job in coaching profession while study material is secondary one. Quality of teaching determines results of your coaching that decides success & future of your coaching while good quality study material is merely a necessary teaching aid that supplements your teaching (video in Hindi)

Image
Read more