| A uniform solid sphere rolls up an inclined plane. |
 |
| Now, choose the correct option. |
A) \[h\propto \frac{1}{m}\]
B) \[h\propto \frac{1}{\tan \theta }\]
C) \[h\propto {{v}^{2}}\]
D) \[h\propto \frac{1}{{{r}^{2}}}\]
Correct Answer: C
Solution :
| Equating rotational KE at bottom with potential energy at top, we have |
| \[\frac{1}{2}m{{\upsilon }^{2}}+\frac{1}{2}I{{\omega }^{2}}=mgh\] |
| \[\frac{1}{2}m{{\upsilon }^{2}}+\frac{1}{2}\times \frac{2}{5}m{{r}^{2}}\times \frac{{{\upsilon }^{r}}}{{{r}^{2}}}=mgh\] |
| \[{{\upsilon }^{2}}\times \left( \frac{1}{2}+\frac{1}{5} \right)=gh\] |
| So, h is independent of mass of ball or angle of inclination. |
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