Question

A hemispherical bowl of radius unity is filled up with water upto the depth $\frac{1}{2}$. The volume of water in the bowl is

• A. $\frac{27\pi }{24}$
• B. $\frac{5\pi }{24}$
• C. $\frac{3\pi }{4}$
• D. None of these

Answer 1

Answer: (B)

$\frac{5\pi }{24}$

In $△OAB$,
$OA=1,OB=\frac{1}{2}$
$\therefore AB=\sqrt{{OA}^2-{OB}^2}=\sqrt{1-\frac{1}{4}}=\frac{\sqrt{3}}{2}$
$\therefore$ Required shaded volume
$=\frac{\pi h}{6}(3a^2+h^2)$
Here $a=AB=\frac{\sqrt{3}}{2},h=\frac{1}{2}$
$\therefore V=\frac{\pi }{6}\times \frac{1}{2}[3{\left(\frac{\sqrt{3}}{2}\right)}^2+{\left(\frac{1}{2}\right)}^2]=\frac{5\pi }{24}$