System Of Particles And Rotational Motion

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Question
CBSEENPH11020654

straight A space force space straight F space equals space straight alpha stack space straight i with hat on top space plus space 3 space straight j with hat on top space plus space 6 space straight k with hat on top space is space acting space at space straight a space point space straight r equals space 2 straight i with hat on top space minus space 6 space straight j with hat on top space minus 12 space straight k with hat on top. The value of α for which angular momentum about origin is conserved is
  • -1

  • 2

  • zero

  • 1

Solution

A.

-1

When the resultant external torque acting on a system is zero, the total angular momentum of a system is zero, the total angular momentum of a system remains constant.This is the principle of the conservation of angular momentum.
Given,
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Question
CBSEENPH11020737

A bar magnet having a magnetic moment of 2 x 104 JT-1 is free to rotate in a horizontal plane. A horizontal magnetic field B = 6 x 10-4 T exists in the space. The work  is done in taking the magnet slowly from a direction parallel to the field to a direction 60o from the field is

  • 0.6 J

  • 12 J

  • 6 J

  • 2 J

Solution

C.

6 J

The work done in rotating a magnetic dipole against the torque acting on it, when placed in a magnetic field is stored inside it in the form of potential energy. 
When magnetic dipole is rotated from initial position θ = θ1 to final position θ = θ2, then work done = MB (cos θ1 = cos θ2)
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Question
CBSEENPH11020641

A body of mass M hits normally a rigid wall with velocity v and bounces back with the same velocity. The impulse experienced by the body is 

  • 1.5 Mv

  • 2 Mv

  • zero

  • Mv

Solution

B.

2 Mv

Impulse |J| = |Δp|
= Mv - (- Mv)
 = 2 Mv

Question
CBSEENPH11020680

A circular disk of moment of inertia It is rotating in a horizontal plane, about its symmetry axis, with a constant angular speed.straight omega subscript straight i Another disk of moment of inertia Ib is dropped coaxially onto the rotating disk. Initially, the second disk has zero angular speed. Eventually, both the disks rotate with constant angular speed.straight omega subscript straight fThe energy lost by the initially rotating disc due to friction is 

  • 1 half fraction numerator straight I subscript straight b superscript 2 over denominator left parenthesis straight I subscript straight t plus straight I subscript straight b right parenthesis end fraction space straight omega subscript straight i superscript 2
  • 1 half fraction numerator straight I subscript straight t superscript 2 over denominator left parenthesis straight I subscript straight t plus straight I subscript straight b right parenthesis end fraction straight omega subscript straight i superscript 2
  • 1 half fraction numerator straight I subscript straight b minus straight I subscript straight t over denominator left parenthesis straight I subscript straight t plus straight I subscript straight b right parenthesis end fraction straight omega subscript straight i superscript 2
  • 1 half fraction numerator straight I subscript straight b straight I subscript straight t over denominator left parenthesis straight I subscript straight t plus straight I subscript straight b right parenthesis end fraction straight omega subscript straight i superscript 2

Solution

D.

1 half fraction numerator straight I subscript straight b straight I subscript straight t over denominator left parenthesis straight I subscript straight t plus straight I subscript straight b right parenthesis end fraction straight omega subscript straight i superscript 2

Loss of energy, 
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Question
CBSEENPH11020551

A circular platform is mounted on the frictionless vertical axle. Its radius R =2 m and its moment of inertia about the axle is 200 kg m2. It is initially at rest. A 50 kg man stands on the edge of the platform and begins to walk along the edge at the speed of 1 ms-1 relative to the ground. Time taken by the man to complete one revolution is

  • π sec

  • 3π/2 sec

  • 2π sec

  • π/2 sec

Solution

A.

π sec

For conservation of angular momentum
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