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`        A force F=-yi+xj acts on the rim of a rigid disc which can freely rotate about z axis along anticlockwise direction in x y plane .Center of disc is at origin .If the work done by the force is 32π in one complete revolution , find the radius of the disc?`
10 months ago

Aman
29 Points
```							As we know that the work done by the torque is Torque.angle rotated. So tor que about centre by applied force is. r*F. That is. Let r be radius (r j cap) *(force). Now doing vector product.  r j^  * (-yi^+xj^).  After solving we get. -ry k^ and work done in complete revolution is. (-ry k^) • 2pi -k^=32pi. The we get final answer r = 16/y.Please report if any correction
```
10 months ago
Utkarsh shukla
12 Points
```							Thanks dude I got my ans 4 let r=xi+yj and then use x^2+y^2 =r^2.        Torq=k(x^2+y^2)=kr^2Work=torq×2π=32πr=4
```
10 months ago
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## COUPON CODE: SELF10

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### Course Features

• 110 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions