A water tank has the shape of an inverted right circular cone with its axis vertical and vertex lowermost. Its semi-vertical angle is `tan^(-1)(0. 5)`. Water is poured into it at a constant rate of 5 cubic metre per hour. Find the rate at which the level of the water is rising at the instant when the depth of water in the tank is 4m.

Explanation

To solve the problem, we need to find the rate at which the level of water is rising in an inverted right circular cone when the depth of water is 4 meters. Here are the steps to arrive at the solution:

### Step 1: Understand the Geometry of the Cone
The cone has a semi-vertical angle given by \( \theta = \tan^{-1}(0.5) \). This means that:
\[ \tan \theta = 0.5 \]
From the definition of tangent, we can relate the radius \( r \) of the water surface to the height \( h \) of the water:
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