A solid object having uniform density  and varying cross-sectional area (which is circular) is placed on smooth horizontal surface as shown. Then what is the value of $\frac{R_2}{R_1}$,if the stress throughout its height remains same and having value equal to c

Stress at any point at distance h from the top is due to the weight of part above it
lets consider a 'h' height of the object, where the radius is 'r'.

 $\frac{\mathrm{mg}}{{\mathrm{\pi r}}^2}=\mathrm{c}\left(\mathrm{constant}\right)$

$m=\frac{c\pi r^2}{g}$

On differentiating m w.r.t r 

we get,$\Rightarrow \mathrm{dm}=\frac{\mathrm{\pi c}2\mathrm{rdr}}{\mathrm{g}}$.....(1)

And the mass of small elemental cylinder is dm

$\mathrm{dm}={\mathrm{\rho \pi r}}^2\mathrm{dh}$.....(2)

From (1) and (2).

we get, ${\mathrm{\rho \pi r}}^2\mathrm{dh}=\frac{2\mathrm{\pi c}}{\mathrm{g}}\mathrm{rdr}$.

${\int }_0^{\mathrm{H}}\mathrm{dh}=\frac{2\mathrm{c}}{\mathrm{\rho g}}{\int }_{{\mathrm{R}}_1}^{{\mathrm{R}}_2}\frac{\mathrm{dr}}{\mathrm{r}}\Rightarrow \mathrm{H}=\frac{2\mathrm{c}}{\mathrm{\rho g}}\ln \left(\frac{{\mathrm{R}}_2}{{\mathrm{R}}_1}\right)$

So, $\frac{{\mathrm{R}}_2}{{\mathrm{R}}_1}={\mathrm{e}}^{\frac{\mathrm{\rho gH}}{2\mathrm{c}}}$