(d) 4
Let ADE be the remaining portion of the cone. Then, from similar Δs AFE and AGC
\(\frac{AF}{AG}=\frac{FE}{GC}\) ⇒ FE = \(\frac{AF.GC}{AG}\)
⇒ r1 = \(\frac{hr}{8}\)
where r1 = radius of cone ADE,
h = height of cone ADE and
r = radius of filled cone ABC
Now, volume of smaller cone
ADE = \(\frac13πr^2_1h = \frac13π\big(\frac{hr}{8}\big)^2.h\)
= \(\frac13\timesπ\times\frac{h^3r^2}{64}\)
Volume of bigger cone ABC = \(\frac13πr^2\times8\)
Given, \(\frac13\timesπ\times\frac{h^3r^2}{64}=12.5\)% of \(\big(\frac83πr^2\big)\)
⇒ \(\fracπ3\times\frac{h^3r^2}{64}=\frac{125}{1000}\times\frac83πr^2\) ⇒ h3 = 64 ⇒ h = 4.