Question

An open cylinder has a round drain with radius $r$ in the bottom of tank. $t=\frac{R^2}{r^2}\sqrt{\frac{2h}{g}}$ governs the time required to empty the tank where $R$ is the radius of the tank and $h$ is the depth of water in cylinder and $g=980cm/s^2$ is the acceleration due to gravity. Find the time required to empty the tank if the radius of the tank is $w$ m and depth of water is $4$m and drain has a radius of $5cm$.

• A. $0.03w^2$
• B. $0.05w^2$
• C. $0.08w^2$
• D. $0.1w^2$

Answer 1

The correct option is A $0.03w^2$

$t=\frac{R^2}{r^2}\sqrt{\frac{2h}{g}}$
$r=5$ cm
$h=4$ m $=400$ cm

$g=980cm/s^2$
Substitute all the values we get,

$t=\frac{w^2}{5^2}\sqrt{\frac{2\times 400}{980}}$

$t=\frac{w^2}{5^2}\sqrt{\frac{2\times 400}{980}}$

$t=0.03w^2$