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PHYSICS-Al, , So Late, , ©. Assertion. Value of radius of gyration of a body, depends on axis of rotation., , Reason, Radius of gyration is root mean etary, distance of particles of the body from the ae :, rotation, [AIIM, , su. Assertion. A solid sphere is rolling on a rough, horizontal surface. Acceleration of contact point is, zero,, , Reason. A solid sphere can roll on the smooth, , IIMS 17], , surface., , Answers and Explanations, , 1, () As there is no external force acting on the, system, the centre of mass continues to remain at rest., , 2. (”) The centre of mass remains at the centre of the, ladder. Hence Fig. (a) represents the correct trace of CM., , 3. () Pattern A is more sturdy, Here the moment, of tension about the fulcurm required to, counterbalance the moment of mg acting at CM of the, rod is maximum., , 4. (In an orbital motion, the direction of angular, momentum vector is perpendicular to the orbital, plane., , 5. (/) If we wrap the right hand around the axis of, rotation with the fingers pointing in the direction of, rotation, then the thumb points in the direction of, angular velocity., , 6. () The motion of planets around the sun is an, example of the conservation of angular momentum., , 7. (“) Angular momentum is conserved only in the, absence of external torque., , 8. (1) pe al, dt, aise MK, Riva ds, 9. (b) About the initial position, the angular, , momentum of the particle is in clockwise sense, Also, the torque due to its weight mg is in clockwise sense., Hence the angular momentum increases., , x), mv, , i, , =|. Assertion. Torque on a body can be zerg eveng, there is a net force on it., Reason. Torque and force on a body are atv, perpendicular. ‘ie io, 59 Assertion. If a satellite is orbiting aroung, planet, then its angular momentum is conseryeg, Reason. Linear momentum conservation feds, , angular momentum conservation., , 10. (1) Radius of gyration of a body depends on the, axis rotation., , 11. (0) Rotational energy is minimum when My Hs, minimum about the axis of rotation. If this Point is at, distance x from 0.3 kg mass, then, , 1 =0.3x? +0.7 (1.4-x), dl _, , Now —=0, dx, or 0.6x + 1.4 (1.4—x)(-1)=0, or x = 0.98 m., , 12. (7) MI. of a rod about a perpendicular axis, through its one end, , , , _ ME, 3, L/2, Total M.I. of two halves, about the axis through O will be ‘, 1=2( 4 )ciay ME, | Oana, 2 118 12, 2, 13: (a) L= lo=2 MR2. 2% — 48MR, 5 Dy 5T, , 14. (b) By conservation of angular momentum,, L= Im= constant, , As the liquid is dropped, it starts spreading out., The moment of inertia increases and angular velocity, decreases. As the liquid starts falling, the moment af, inertia again decreases and angular velocity increases:, , 15. () The M.L of B about its own length is 22, We need to find MI. of Aand C about a perpendicular, axis through their centre., , 2 2 2