A solid sphere and a solid cylinder of same mass are rolled down on two inclined planes of heights h₁ & h₂. If at the bottom of the plane of two objects have same linear velocities, then ratio of  h₁ to h₂  is

Since the mechanical energy is conserved

$\mathrm{mgh} = {\mathrm{k}}_{\mathrm{R}}+{\mathrm{k}}_{\mathrm{T}}$

From the question in case of pure rolling,

$\frac{{\mathrm{k}}_{\mathrm{R}}}{{\mathrm{k}}_{\mathrm{T}}}=\frac{1}{2}$for a solid cylinder and

$\frac{{\mathrm{k}}_{\mathrm{R}}}{{\mathrm{k}}_{\mathrm{T}}}=\frac{2}{5}$ for a solid spherical.
Both have the same linear velocities, or translational kinetic energies, at the bottom.

$\frac{5}{7}mg{h}_1=\frac{2}{3}mg{h}_2$

$\frac{h_1}{h_2}=\frac{14}{15}$

Hence the correct answer is 14:15.