A cylindrical tank of radius R and height H is half-filled with water. A solid metal cone of base radius frac R 2 and height H is completely immersed in the tank with its base facing down. By what fraction of H does the water level rise?

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EEJ MAIN Aptitude QUESTION #10997

Question 1

A cylindrical tank of radius $R$ and height $H$ is half-filled with water. A solid metal cone of base radius $\frac{R}{2}$ and height $H$ is completely immersed in the tank with its base facing down. By what fraction of $H$ does the water level rise?

$\frac{1}{6}$
$\frac{1}{8}$
$\frac{1}{12}$✔️
$\frac{1}{24}$

Correct Answer Explanation

Volume of cone: $V = \frac{1}{3}\pi (\frac{R}{2})^2 H = \frac{\pi R^2 H}{12}$. The tank's cross-sectional area is $\pi R^2$. Rise in water level: $\Delta h = \frac{V}{\pi R^2} = \frac{H}{12}$. So the water level rises by $\frac{1}{12}$ of $H$.

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