Question

If a sphere rolling on an inclined plane with velocity $v$ without slipping, the vertical height of the incline in terms of velocity will be:

• A. $ \frac{7\upsilon }{10g} $
• B. $ \frac{7{{\upsilon }^{2}}}{10g} $
• C. $ \frac{2{{\upsilon }^{2}}}{5g} $
• D. $ \frac{2\upsilon }{5g} $

Answer 1

Answer: (B)

The velocity of solid sphere on the bottom of inclined plane is
$v=\sqrt{\left[\frac{2 g h}{\left(1+I / M R^{2}\right)}\right]}$
where, $I=$ moment of inertia 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} $
$\therefore v=\sqrt{\left[\frac{2\, g h}{\left(1+\frac{2}{5}\right)}\right]}=\sqrt{\left(\frac{10}{7}\, g h\right)}$
$\therefore h=\frac{7 v^{2}}{10\, g}$