A solid sphere and a hollow cylinder roll up without slipping on same inclined plane with same initial speed $$\upsilon$$. The sphere and the cylinder reaches upto maximum heights $$h_1$$ and $$h_2$$, respectively, above the initial level. The ratio $$h_1 : h_2$$ is $$\frac{n}{10}$$. The value of n is _____.

A solid sphere and a hollow cylinder roll up an incline with the same initial velocity $$v$$. The ratio of the heights they reach is $$\frac{h_1}{h_2} = \frac{n}{10}$$. We need to find $$n$$.

Apply energy conservation for rolling bodies.

For a body rolling without slipping, the total kinetic energy converts to potential energy at the maximum height:

$$ \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 = mgh $$

Using $$\omega = v/r$$ (rolling condition): $$\frac{1}{2}mv^2\left(1 + \frac{I}{mr^2}\right) = mgh$$

$$ h = \frac{v^2}{2g}\left(1 + \frac{I}{mr^2}\right) $$

Calculate height for the solid sphere.

For a solid sphere: $$I = \frac{2}{5}mr^2$$, so $$\frac{I}{mr^2} = \frac{2}{5}$$

$$ h_1 = \frac{v^2}{2g}\left(1 + \frac{2}{5}\right) = \frac{v^2}{2g} \times \frac{7}{5} = \frac{7v^2}{10g} $$

Calculate height for the hollow cylinder.

For a hollow cylinder (thin-walled): $$I = mr^2$$, so $$\frac{I}{mr^2} = 1$$

$$ h_2 = \frac{v^2}{2g}(1 + 1) = \frac{v^2}{g} $$

Find the ratio.

$$ \frac{h_1}{h_2} = \frac{7v^2/(10g)}{v^2/g} = \frac{7v^2}{10g} \times \frac{g}{v^2} = \frac{7}{10} $$

Therefore $$n = 7$$.

The answer is $$\boxed{7}$$.