Question

A hemispherical bowl fits exactly on the open surface of a right cone with height $5cm$ and radius $3.5cm$. The volume of the solid formed is

[Take $\pi =\frac{22}{7}$]

• A. 144 cm³
• B. 132 cm³
• C. 154 cm³
• D. 169 cm³

Answer 1

The correct option is C

154 cm³

Total volume of the solid formed $=$ Volume of the cone $+$ Volume of the hemisphere

$=\frac{1}{3}\pi r^{2}h+\frac{2}{3}\pi r^3$

$=\frac{1}{3}\pi r^2[h+2r]$

$=\frac{1}{3}\times \pi \times {3.5}^2[5+7]$

$=\frac{1}{3}\times \frac{22}{7}\times 3.5\times 3.5\times 12$

$=22\times 0.5\times 3.5\times 4$

$=154c{m}^3$