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`        A solid sphere of radius R is placed at rest on a rough horizontal surface. The sphere is given a horizontal impulse at height h above its lowest point. Consequently, it starts rolling without slipping. The value of h is`
2 years ago

Vikas TU
8378 Points
```							Moment zero at point of contct.Hence, F*h = I*a/RFh = (2mR^2/5 + mR^2)a/Rh = (7mR/5)a meterwhere a is the accleration and m is the mass of the sphere.
```
2 years ago
ruhi
16 Points
```							according to the question the sphere was is in rest therefore the net torque=0I×h - F×R =0MR^2 ×h =(2/5 )MR^2 ×Rh/R= 2/5this is the required ratio...here R= radius of the sphereI= moment of inertia applied to the sphereF= moment of inertia about diameter
```
9 months 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