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03 - Rotational Motion Ex. 1 - 3 Module-2

LEVEL # 1 Questions Conceptual Problems Q.6 A particle, situated in an object, moves with based on Q.1 A wheel starts rotating from rest and attains an angular velocity of 60 rad/sec in 5 seconds. The total angular displacement in radians will be : (A) 60 (B) 80 (C) 100 (D) 150 Q.2 A body rotates at 300 rotations per minute. The value in radian of the angle described in 1 sec is (A) 5 (B) 5 (C) 10 (D) 10 Q.3 A chain couples and rotates two wheels in a bicycle. The radii of bigger and smaller wheels in a bicycle. The radii of bigger and smaller wheels are 0.5m and 0.1. respectively. The bigger wheel rotates at the rate of 200 rotations per minute, then the rate of rotation of smaller wheel will be - (A) 1000 rpm (B) 50/3 rpm (C) 200 rmp (D) 40 rpm angular acceleration of 6 rad/sec2 and with 2 rad/sec angular velocity. If the radius of the circular path is 1m, its total acceleration in m/sec2 will be : (A) 1 (B) 100 (C) 10 (D) Q.7 A particle starts from rest under
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5- Refraction at curved surface Theory Module-5-1

1.1 PRISM : It is a transparent medium whose refracting surfaces are not parallel but are inclined to each other. 1.2 TERMS RELATED TO PRISM : (A) Angle of Prism or Refracting Angle (A) : It is the angle between the faces on which light is incident and from which it emerges. (B) Angle of Deviation (  ) : It is the angle between the emergent and the incident ray. In other words, it is the angle through which incident ray turns while passing through a prism.  = (i – r1) + (e – r2) = i + e – (r1 + r2) = i + e – A 1.3 CONDITION OF NO EMERGENCE : 1.4 CRITICAL ANGLE : It is the angle of prism above which incident light ray on first surface will not emerge out from the second surface for all possible values of angle of incidence A = 2 C 1.5 CONDITION OF GRAZING EMERGENCE : If a ray can emerge out of a prism , the value of angle of incidence i for which angle of emergence e = 90° is called condition of grazing emergence. i.e., r2 = c  , r1 = A – r2 = A – c (As r1 + r2 = A)
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8-Circular Motion

8.1 (D) The maximum angular speed of the hoop corresponds to the situation when the bead is just about to slide upwards. The free body diagram of the bead is For the bead not to slide upwards. m2 (r sin 45°) cos 45° – mg sin 45° < N (1) where N = mg cos 45° + m2 (r sin 45ยบ) sin 45° (2) From 1 and 2 we get.  = rad / s. 8.2 (C) Let v be the speed of particle at B, just when it is about to loose contact. From application of Newton's second law to the particle normal to the spherical surface. mv2 = mg sin  (1) r Applying conservation of energy as the block moves from A to B.. 1 mv2 = mg (r cos  – r sin ) (2) 2 Solving 1 and 2 we get 3 sin  = 2 cos  8.3 (A) As the mass is at the verge of slipping ∴ mg sin37 –  mg cos37 = m2r 6 – 8 = 4.5   = 3 16 8.4 (B) As when they collide vt  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 8.5 (C) The acceleration vector s
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