Question

A cylindrical tank of height 0.4 m is open at the top and has a diameter 0.16 m. Water is filled in it up to a height of 0.16 m. How long it will take to empty the tank through a hole of radius $5\times10^{-3}$ m at its bottom ?

• A. 46.26 sec.
• B. 4.6 sec.
• C. 462.6 sec.
• D. 0.46 sec.

Answer 1

Answer: (A)

46.26 sec.

Answer 2

The time ($t$) taken to empty a cylindrical tank of radius $R$ from an initial water height $h_0$ through a hole of radius $r_h$ at the bottom is given by the formula:

$t = \frac{R^2}{r_h^2} \sqrt{\frac{2h_0}{g}}$

Where:

• $R$ = radius of the tank = 0.08 m
• $r_h$ = radius of the hole = $5 \times 10^{-3}$ m
• $h_0$ = initial height of water = 0.16 m
• $g$ = acceleration due to gravity = $9.8 \text{ m/s}^2$

Substituting the values:

$t = \frac{(0.08)^2}{(5 \times 10^{-3})^2} \sqrt{\frac{2 \times 0.16}{9.8}}$ $t = \frac{0.0064}{25 \times 10^{-6}} \sqrt{\frac{0.32}{9.8}}$ $t = 256000 \times \sqrt{0.032653}$ $t = 256 \times 0.1807$ $t \approx 46.26 \text{ seconds}$