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`         question 28 of chapter 10 rotation and rolling...///////////////////////////////`
2 years ago

we will so
101 Points
```							 = 100 rev/min = 5/8 rev/s = 10/3 rad/s = 10 rev = 20  rad, r = 0.2 mAfter 10 revolutions the wheel will come to rest by a tangential forceTherefore the angular deacceleration produced by the force =  = 2/2Therefore the torque by which the wheel will come to an rest = Icm ×  F × r = Icm ×   F × 0.2 = 1/2 mr2 × [(10/3)2 / (2 × 20)] F = 1/2 × 10 × 0.2 × 100 2 / (9 × 2 × 20)= 5 / 18 = 15.7/18 = 0.87 N.
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2 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