An inverted conical vessel whose height is 10 cm and the radius of whose base is 5 cm is being filled with water at the uniform rate of `1.5cm^(3)//min`. Find the rate at which the level of water in the vessel is rising when the depth is 4 cm.

Answer

To solve the problem step by step, we will follow these steps:

### Step 1: Understand the problem
We have an inverted conical vessel with a height (H) of 10 cm and a base radius (R) of 5 cm. Water is being poured into the vessel at a rate of 1.5 cm³/min. We need to find the rate at which the water level (h) is rising when the depth of the water is 4 cm.

### Step 2: Write the formula for the volume of a cone
The volume (V) of a cone is given by the formula:
\[ V = \frac{1}{3} \pi r^2 h \]
where \( r \) is the radius of the water surface at height \( h \).

### Step 3: Establish the relationship between r and h
Since the cone is similar at different heights, we can establish a relationship between the radius \( r \) and the height \( h \):
\[ \frac{r}{h} = \frac{R}{H} \]
Substituting the known values:
\[ \frac{r}{h} = \frac{5}{10} = \frac{1}{2} \]
Thus, we can express \( r \) in terms of \( h \):
\[ r = \frac{1}{2} h \]

### Step 4: Substitute r in the volume formula
Now, substitute \( r \) in the volume formula:
\[ V = \frac{1}{3} \pi \left(\frac{1}{2} h\right)^2 h = \frac{1}{3} \pi \left(\frac{1}{4} h^2\right) h = \frac{1}{12} \pi h^3 \]

### Step 5: Differentiate the volume with respect to time
Now we differentiate both sides of the volume equation with respect to time \( t \):
\[ \frac{dV}{dt} = \frac{1}{12} \pi \cdot 3h^2 \frac{dh}{dt} \]
This simplifies to:
\[ \frac{dV}{dt} = \frac{1}{4} \pi h^2 \frac{dh}{dt} \]

### Step 6: Substitute known values
We know that \( \frac{dV}{dt} = 1.5 \) cm³/min and we need to find \( \frac{dh}{dt} \) when \( h = 4 \) cm. Substitute these values into the equation:
\[ 1.5 = \frac{1}{4} \pi (4^2) \frac{dh}{dt} \]
Calculating \( 4^2 \):
\[ 1.5 = \frac{1}{4} \pi (16) \frac{dh}{dt} \]
This simplifies to:
\[ 1.5 = 4\pi \frac{dh}{dt} \]

### Step 7: Solve for \( \frac{dh}{dt} \)
Now, isolate \( \frac{dh}{dt} \):
\[ \frac{dh}{dt} = \frac{1.5}{4\pi} \]
This can be simplified to:
\[ \frac{dh}{dt} = \frac{3}{8\pi} \text{ cm/min} \]

### Final Answer
The rate at which the level of water in the vessel is rising when the depth is 4 cm is:
\[ \frac{3}{8\pi} \text{ cm/min} \]