Physics-DPP-1-14.Rotational Motion

PHYSICS Topic: ROTATIONAL MOTION Date: 1. The angular momentum of a projectile projected at angle θ with, the horizontal with speed u about the point of projection when it is at the highest point is : (a) mu2 sin2 θ g (b) mu cosθ g (c) mu3 sin2 θ cosθ (d) none of these 2g 2. Radius of gyration of a body about an axis at a distance of 4 cm from its centre of mass is 8 cm. The radius of gyration about a parallel axis through its centre of mass is : (a) 7 cm (b) 3 cm (c) 4 cm (d) none of these 3. The distance between the sun and earth be r then the angular momentum of the earth around the sun is proportional to : (a) (b) r3/ 2 (c) r (d) none of these 4. Initial angular velocity of a circular disc of mass M is ω1 . Then two small spheres of mass m are attached gently to two diametrically opposite points on the edge of the disc. What is the final angular velocity of the disc ? (a) æçM + çè m ö÷ ÷ 1 (b) æçM + ç m ö÷ ÷ 1 (c) æç M ö÷ (d) ç M ö÷ M ø è m ø èM + 4mø èM + 2mø 5. If a body completes a vertical circle, then : (a) total energy of body remains constant (b) angular momentum remains constant (c) angular velocity remains constant (d) none of the above 6. A circular disc X of radius R is made from an iron plate of thickness t, and another disc Y of radius 4 R is made from an iron plate of thickness t / 4. Then the relation between the moment of inertia IX and IY is : (a) IY = 32 IX (b) IY = 16 IX (c) IY = IX (d) IY = 64 IX 7. A particle performing uniform circular motion has angular momentum L. If its angular frequency is doubled and its kinetic energy halved, then the new angular momentum is : (a) L L 4 (b) 2 L (c) 4 L (d) 2 8. The moment of inertia of uniform semicircular disc of mass M and radius r about a line perpendicular to the plane of the disc through the centre is : (a) 1 Mr 2 4 (b) 2 Mr 2 5 (c) Mr2 (d) 1 Mr 2 2 9. The instantaneous velocity of point B of the given rod of length 0.5 m is 3 m/s in the represented direction. The angular velocity of the rod for minimum velocity of end A is : (a) 1.5 rad/s (b) 5.2 rad/s (c) 2.5 rad /s (d) none of these 10. A nearly massless rod is pivoted at one end so that it can swing freely as a pendulum. Two masses, 2 m and m, are attached to it at distance b and 3b, respectively, from the pivot. The rod is held horizontal and then released. Find its angular acceleration at the instant it released. (a) 2g / 11b (b) 4g/ 17 b (c) 5g / 11b (d) 5g / 12 b PHYSICS Topic: ROTATIONAL MOTION Date: Answer Key 1. (c) 2. (a) 3. (a) 4. (c) 5. (d) 6. (d) 7. (a) 8. (d) 9. (b) 10. (c)

Comments

Post a Comment

Popular posts from this blog

PHYSICS-15-10- 11th (PQRS) SOLUTION

XI (PQRS) PHYSICS REVIEW TEST-5 Select the correct alternative. (Only one is correct) [15 × 3 = 45] There is NEGATIVE marking. 1 mark will be deducted for each wrong answer. Q.31 Two trains, which are moving along different tracks in opposite directions towards each other, are put on the same track due to a mistake. Their drivers, on noticing the mistake, start slowing down the trains when the trains are 300 m apart. Graphs given below show their velocities as function of time as the trains slow down. The separation between the trains when both have stopped, is: (A) 120 m (B) 280 m (C) 60 m (D*) 20 m 1 [Sol: x1 = 2 1 × 10 × 40 = 200 m; x2 = – 2 × 8 × 20 = –80 m  xreletive = x1 + x2 = 280  separation = 300 – 280 = 20 m (D) ] Q.32 One end of a thin uniform rod of length L and mass M is riveted to the centre of a uniform circular disc of radius r and mass 2 M so that the rod is normal to the disc. The centre of mass of the combination from the centre of the disc is at dist
Read more

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 45º) 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
Read more

4. THEORY-Current Electrictricity

CURRENT ELECTRICITY  1 . ELECTRIC CURRENT (a) Time rate of flow of charge through a cross sectional area is called Current. if q charge flows in time interval t then average current is given by q Iav = t and Instantaneous current i =. lim q  dq t0 t dt (b) Direction of current is along the direction of flow of positive charge or opposite to the direction of flow of negative charge. But the current is a scalar quantity. i i q+ velocity q– velocity SI unit of current is ampere and 1 Ampere = 1 coloumb/sec 1 coloumb/sec = 1A 2 . CONDUCTOR In some materials, the outer electrons of each atom or molecule are only weakly bound to it. These electrons are almost free to move throughout the body of the material and are called free electrons. They are also known as conduction electrons. When such a material is placed in an electric field, the free electrons drift in a direction opposite to the field. Such materials are called conductors. 3 . INSULATOR Another class o
Read more