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The speed of a homogenous solid sphere after rolling down an inclined plane of vertical height h from rest without sliding is

1. sqrt(10 / 7) gh
2. sqrt(gh)
3. sqrt(6 / 5) gh
4. sqrt(4 / 3) gh

Correct answer: sqrt(10 / 7) gh

Solution

For a rolling object, the total energy is conserved. The potential energy at height h is converted into translational and rotational kinetic energy. Using the moment of inertia of a solid sphere (I = 2/5 mR²) and solving for velocity, we get v = sqrt((10/7)gh).

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