A thin uniform spherical shell and a uniform solid cylinder of the same mass and radius are allowed to roll down a fixed incline without slipping, starting from rest. The ratio $\left(\frac{t_{\text{sph}}}{t_{\text{cyl}}}\right)$  of times taken by them to roll down the same distance is

From the free body diagram it is clear that we get the equation $mg\sin \text{θ}-f=ma\mathrm{.......}\left(1\right)$
Torque  $\tau =fr=I\alpha$
$\Rightarrow fr=I.\frac{a}{r}\mathrm{.....}\left(2\right)\left(\because a=r\alpha ,\text{pure\hspace{0.17em}rolling}\right)$  
Solving 1 and 2
$\Rightarrow mg\sin \text{θ}-\frac{Ia}{r^2}=ma \left(\because I=m{k}^2\right)$

$\Rightarrow mg\sin \text{θ}-\frac{m{k}^{2}a}{r^2}=ma$

$\Rightarrow g\sin \text{θ}=a\left(1+\frac{k^2}{r^2}\right)$

$\Rightarrow a=\frac{g\sin \text{θ}}{1+\frac{k^2}{r^2}}$

Time taken by the bodies to roll down a distance ‘l’ is

$t=\sqrt{\frac{2l}{a}}=\sqrt{\frac{2l\left(1+\frac{k^2}{R^2}\right)}{g\sin \text{θ}}}$

For shell :   $\frac{k^2}{r^2}=\frac{2}{3}$
For cylinder :  $\frac{k^2}{r^2}=\frac{1}{2}$
$\frac{t_{\text{sph}}}{t_{cy}}=\sqrt{\frac{1+\frac{2}{3}}{1+\frac{1}{2}}}=\sqrt{\frac{5\times 2}{3\times 3}}=\frac{\sqrt{10}}{3}$