Question

please solve this question. immediately required solution

Answer 1

C1:   Solid right circular cone of height H, base radius R.
          Volume = V1 : 1/3 π R² H

C2:   solid right circular cone of height h, base radius R.
          Volume = V2 = 1/3 π R² h

C3:   hollow big cone after removal of smaller cone
         Volume = V1 - V2 = 1/3 π R² (H - h)

As density is constant, mass is proportional to the volume.

    Since the cone is triangular, and symmetric about vertical axis through its apex, the center of mass is at 1/4 th height from the base. It is on the axis.

    COM1 of C1: H/4 from base.
    COM2 of C2:  h/4 from base.
    COM3 of C3 :  h from base     =  given

=>  h * 1/3 π R² (H - h) + h/4 * 1/3 π R² h = H/4 * 1/3 π R² H
=>  h (H - h) + h² / 4 = H² / 4
=>  h H - 3h²/4 = H²/4
=>  3 h² - 4 H h + H² = 0
=>  (3h - H) (h - H) = 0
=>   h = H/3         (as  h = H means entire solid cone is removed).

Answer 2

volume of cylinder :- 1/3 π R2 h