A sphere rolls down on an inclined plane of inclination $\theta$. What is the acceleration as the sphere…
A sphere rolls down on an inclined plane of inclination $\theta$. What is the acceleration as the sphere reaches the bottom?
- $\frac{5}{7} g \sin \theta$
- $\frac{3}{5} g \sin \theta$
- $\frac{2}{7} g \sin \theta$
- $\frac{2}{5} g \sin \theta$
Solution
$a=\frac{g \sin \theta}{1+\frac{K^2}{R^2}}=\frac{g \sin \theta}{1+\frac{2}{5}}=\frac{5}{7} g \sin \theta$
Asked in: TEST SERIES MHT-CET Full Test 6
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