An inverted conical flask is being filled with water at the rate of `3cm^(3)`/sec. The height of the flask is 10cm and the radius of the base is 5cm. How fast is the water level rising when the level is 4cm?

Options

A. `(4)/(3)pi` cm/sec
B. `(3)/(4pi)` cm/sec
C. `(3pi)/(4)` cm/sec
D. `(4)/(3pi)` cm/sec

Answer

To solve the problem step by step, we need to find how fast the water level is rising in the inverted conical flask when the water level is at 4 cm.

### Step 1: Understand the Geometry of the Cone
The flask is an inverted cone with:
- Height (H) = 10 cm
- Radius (R) = 5 cm

### Step 2: Relate the Radius and Height of the Water
Let:
- h = height of the water in the cone (in cm)
- r = radius of the water surface at height h (in cm)

Since the cone is similar in shape regardless of the height of the water, we can use the ratio of the dimensions:
\[ \frac{r}{h} = \frac{R}{H} = \frac{5}{10} = \frac{1}{2} \]
Thus, we can express r in terms of h:
\[ r = \frac{1}{2}h \]

### Step 3: Write the Volume of Water in the Cone
The volume (V) of water in the cone is given by the formula:
\[ V = \frac{1}{3} \pi r^2 h \]
Substituting \( r = \frac{1}{2}h \):
\[ V = \frac{1}{3} \pi \left(\frac{1}{2}h\right)^2 h = \frac{1}{3} \pi \frac{1}{4} h^2 h = \frac{1}{12} \pi h^3 \]

### Step 4: Differentiate the Volume with Respect to Time
We know the volume is changing with respect to time as water is being added:
\[ \frac{dV}{dt} = \frac{1}{12} \pi \frac{d(h^3)}{dt} \]
Using the chain rule:
\[ \frac{d(h^3)}{dt} = 3h^2 \frac{dh}{dt} \]
Thus:
\[ \frac{dV}{dt} = \frac{1}{12} \pi (3h^2 \frac{dh}{dt}) = \frac{1}{4} \pi h^2 \frac{dh}{dt} \]

### Step 5: Substitute Known Values
We know the rate at which the volume is increasing:
\[ \frac{dV}{dt} = 3 \text{ cm}^3/\text{s} \]
Now we need to find \(\frac{dh}{dt}\) when \(h = 4 \text{ cm}\):
\[ 3 = \frac{1}{4} \pi (4^2) \frac{dh}{dt} \]
Calculating \(4^2\):
\[ 3 = \frac{1}{4} \pi (16) \frac{dh}{dt} \]
Simplifying:
\[ 3 = 4\pi \frac{dh}{dt} \]
Thus:
\[ \frac{dh}{dt} = \frac{3}{4\pi} \text{ cm/s} \]

### Final Answer
The water level is rising at a rate of:
\[ \frac{dh}{dt} = \frac{3}{4\pi} \text{ cm/s} \]