Coffee is coming out from a conical filter, with height and diameter both are 15 cm into a cylindrical coffee pot with a diameter 15 cm. The rate at which coffee comes out from the filter into the pot is 100 cu cm/min.
The rate (in cm/min) at which the level in the pot is rising at the instance when the coffe in the pot is 10 cm, is

Options

A. `(9)/(16pi)`
B. `(25)/(9pi)`
C. `(5)/(3pi)`
D. `(16)/(9pi)`

Answer

To solve the problem step by step, we will use the relationship between the volume of the cylindrical pot and the rate at which coffee is being poured into it.

### Step 1: Identify the dimensions of the cylindrical pot
The diameter of the cylindrical pot is given as 15 cm. Therefore, the radius \( r \) of the pot is:
\[ r = \frac{15}{2} = 7.5 \text{ cm} \]

### Step 2: Write the formula for the volume of the cylinder
The volume \( V \) of the cylindrical pot can be expressed as:
\[ V = \pi r^2 h \]
where \( h \) is the height of the coffee in the pot.

### Step 3: Differentiate the volume with respect to time
To find the rate at which the height \( h \) is changing, we differentiate the volume with respect to time \( t \):
\[ \frac{dV}{dt} = \pi r^2 \frac{dh}{dt} \]
Since the radius \( r \) is constant, we can factor it out.

### Step 4: Substitute the known values
We know that coffee is coming out at a rate of \( \frac{dV}{dt} = 100 \text{ cm}^3/\text{min} \). Substituting the radius into the differentiated volume equation:
\[ 100 = \pi (7.5)^2 \frac{dh}{dt} \]
Calculating \( (7.5)^2 \):
\[ (7.5)^2 = 56.25 \]
Thus, we have:
\[ 100 = \pi (56.25) \frac{dh}{dt} \]

### Step 5: Solve for \( \frac{dh}{dt} \)
Now, we can solve for \( \frac{dh}{dt} \):
\[ \frac{dh}{dt} = \frac{100}{\pi \times 56.25} \]

### Step 6: Simplify the expression
Calculating \( \pi \times 56.25 \):
\[ \pi \times 56.25 \approx 176.71 \text{ (using } \pi \approx 3.14\text{)} \]
Thus:
\[ \frac{dh}{dt} \approx \frac{100}{176.71} \approx 0.566 \text{ cm/min} \]

### Step 7: Final Calculation
To get a more precise value, we can use the exact value of \( \pi \):
\[ \frac{dh}{dt} = \frac{100}{56.25\pi} = \frac{100}{176.71} \approx 0.566 \text{ cm/min} \]

### Conclusion
The rate at which the level in the pot is rising when the coffee in the pot is 10 cm is approximately:
\[ \frac{dh}{dt} \approx 0.566 \text{ cm/min} \]