MCQ
A solid sphere and a disc of same mass and radius starts rolling down a rough inclined plane, from the same height the ratio of the time taken in the two cases is
- A$15:14$
- B$\sqrt {15} :\sqrt {14} $
- C$14 : 15$
- ✓$\sqrt {14} :\sqrt {15} $
✓
Answer
Correct option: D.
$\sqrt {14} :\sqrt {15} $
d
Time of descent $t = \frac{1}{{\sin \theta }}\sqrt {\frac{{2h}}{g}\left( {1 + \frac{{{k^2}}}{{{R^2}}}} \right)} $
Time of descent $t = \frac{1}{{\sin \theta }}\sqrt {\frac{{2h}}{g}\left( {1 + \frac{{{k^2}}}{{{R^2}}}} \right)} $
= $\sqrt {\frac{{{{\left( {1 + \frac{{{k^2}}}{{{R^2}}}} \right)}_{{\rm{sphere}}}}}}{{{{\left( {1 + \frac{{{k^2}}}{{{R^2}}}} \right)}_{{\rm{disc}}}}}}} $
$ = \sqrt {\frac{{1 + \frac{2}{5}}}{{1 + \frac{1}{2}}}} $$ = \sqrt {\frac{7}{5} \times \frac{2}{3}} = \sqrt {\frac{{14}}{{15}}} $
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