Answer to Question #151706 in Molecular Physics | Thermodynamics for Tana

Question

A conical shaped tank of height H and width W is placed above a spherical water tank of radius R. Water drains from the conical tank into the spherical tank through a small opening of radius r. Similarly, the spherical tank empties though a small opening of radius r located at its bottom. Find the time it would take to drain both the tanks.

Answer 1

Height of Cone = H

Width of cone = W

Width of cone = Circumference of cone = 2πR'

Base Radius of cone = W/2π

Volume of cone = 1/3πr²h

Volume of cone = $\frac{1}{3}×π×(\frac{W}{2π})^2×H = \frac{W²H}{12}$

Volume of spherical tank = 4/3πR³

Total volume of both objects = $\dfrac{W²H}{12π} + \dfrac{4πR³}{3} =$

$\dfrac{W²H + 16πR⁴}{12π}$

Since the radii of the openings of both the spherical and the conical tanks are equal, their rates are equal

The rate at which water moves away from the tanks = 2πr m³/sec

The time taken = volume/rate = $\dfrac{W²H + 16πR⁴}{12π} × \dfrac{1}{2πr} =$

$\dfrac{W²H + 16πR⁴}{24πr}$