Question

A vessel is in the form of an inverted cone is filled with water to the brim its height is 20cm and diameter is 16.8cm. Two equal solid cones are dropped in it so that they are fully submerged. As a result one third of the water in the original cone over flows. What is the volume of each of the solid cones submerged?

Answer 1

$\large{\underline{\rm{\blue{\bf{Question:-}}}}}$

A vessel in the form of an inverted cone is filled with water to the brim. Its height is 20 cm and diameter is 16.8 cm. Two equal solid cones are dropped in it so that they are fully submerged. As a result, one-third of the water in the original cone overflows. What is the volume of each of the solid cone submerged?

$\large{\underline{\rm{\blue{\bf{Given:-}}}}}$

Height of the conical vessel = 20 cm

Diameter = 16.8 cm

$\large{\underline{\rm{\blue{\bf{To \: Find:-}}}}}$

The volume of each of the solid cones submerged.

$\large{\underline{\rm{\blue{\bf{Solution:-}}}}}$

We know that,

• R = Radius
• H = Height
• d = Diameter

Given that,

Height of the conical vessel (h) = 20 cm

Diameter = 16.8 cm

Then, radius (r) = $\sf \dfrac{Diameter}{2}$

Radius (r) = $\sf \dfrac{16.8}{2}$

$\sf \longrightarrow 8.4 \: cm$

Volume of water filled in it,

$\sf \implies \dfrac{1}{3} \pi \: r^{2}h$

Substituting them, we get

$\sf \implies \dfrac{1}{3} \pi \times 8.4 \times 8.4 \times 20 \: cm^{3}$

$\boxed{\sf Value \: of \: \pi =\sf \dfrac{22}{7} }$

$\sf \implies \dfrac{1}{3} \times \dfrac{22}{7} \times 8,4 \times 8.4 \times 20 \: cm^{3}$

$\sf \implies 1478.4 \: cm^{3}$

Therefore,

$\sf \dfrac{1}{3} \%$ volume of water = $\sf 1478.4 \times \dfrac{1}{3}$

$\sf \implies 492.8 \: cm^{3}$

∴ Volume of two equal solid cones = $\sf 492.8 \: cm^{3}$

And, volume of one cone = $\sf \dfrac{Total \: volume \: of \: two \: cones}{2}$

$\sf \implies \dfrac{492.8}{2}$

$\sf = \underline{\underline{246.4 \: cm^{3}}}$

Therefore, the volume of each of the solid cones submerged is 246.4 cm³