\(\sqrt{\frac{2 l\left(1+\frac{K^{2}}{R^{2}}\right)}{g \sin \theta}}\)

Here, / is length of incline plane

For solid cylinder \(K^{2}=\frac{R^{2}}{2}\)

For hollow cylinder \(K^{2}=R^{2}\)

Hence, solid cylinder will reach the bottom first.