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A smooth ring of mass M can slide on a fixed horizontal rod.A massless string tied to the ring passes over a fixed smooth Pulley of mass M and carries a block of mass 2m.At a instant the string between ring and Pulley makes an angle 30° with horizontal.ring acceleration of block and ring,normal force due to ring on the rod,reaction force on the Pulley due to support
A smooth ring of mass M can slide on a fixed horizontal rod.A massless string tied to the ring passes over a fixed smooth Pulley of mass M and carries a block of mass 2m.At a instant the string between ring and Pulley makes an angle 30° with horizontal.ring acceleration of block and ring,normal force due to ring on the rod,reaction force on the Pulley due to support

```
3 years ago

Ankit Raj
39 Points
```							I must say that image will explain more clearly but no problem. For 2M mass 2Mg-T=2Ma ,for ring Tcos30°=Ma` one equation from constraint relationship that length of string remains same so acceleration of both ends will be same therefore a`cos30°=a Answers would be accel.of 2M (3/5)g accel. Of ring (2√3/5)g  Tension (4/5)Mg. Net supporting force on pulley √3T as there will be a horizontal as well.Normal reaxn on ring (2/5)Mg
```
3 years ago
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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