A solid cone of base radius 10 cm is cut into two parts

Let the height of the cone be H.
Now, the cone is divided into two parts by the parallel plane

∴ OC = CAH2

Now, In ∆OCD and OAB

∠OCD = OAB          (Corresponding angles)

∠ODC = OBA          (Corresponding angles)

By AA-similarity criterion ∆OCD ∼ ∆OAB

$\therefore \frac{\mathrm{CD}}{\mathrm{AB}}=\frac{\mathrm{OC}}{\mathrm{OA}}$

$\Rightarrow \frac{\mathrm{CD}}{10}=\frac{H}{2 \times H}$

$\Rightarrow \mathrm{CD}=5 \mathrm{~cm}$

$\frac{\text { Volume of first part }}{\text { Volume of second part }}=\frac{\frac{1}{3} \pi(\mathrm{CD})^{2}(\mathrm{OC})}{\frac{1}{3} \pi \mathrm{CA}\left[(\mathrm{AB})^{2}+(\mathrm{AB})(\mathrm{CD})+\mathrm{CD}^{2}\right]}$

$=\frac{(5)^{2}}{\left[(10)^{2}+(10)(5)+5^{2}\right]}$

$=\frac{25}{100+50+25}$

$=\frac{25}{175}$

$=\frac{1}{7}$