A) \[\frac{7\upsilon }{10g}\]
B) \[\frac{7{{\upsilon }^{2}}}{10g}\]
C) \[\frac{2{{\upsilon }^{2}}}{5g}\]
D) \[\frac{2\upsilon }{5g}\]
Correct Answer: B
Solution :
The velocity of solid sphere on the bottom of inclined plane is \[\upsilon =\sqrt{\left[ \frac{2gh}{{{(1+I/MR)}^{2}}} \right]}\] where, \[I=M.I\] of sphere, \[M=\] mass of sphere, and \[R=\]radius of sphere The moment of inertia of solid sphere about its diameter \[I=\frac{2}{5}M{{R}^{2}}\] \[\therefore \] \[\upsilon =\sqrt{\left[ \frac{2gh}{\left( 1+\frac{2}{5} \right)} \right]}=\sqrt{\left( \frac{10}{7}gh \right)}\] \[\therefore \] \[h=\frac{{{7}_{{{\upsilon }^{2}}}}}{10g}\]
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