Water is flowing into a right circular conical vessel, 45cm deep and 27cm in diameter at the rate of `11 cm^(3)//"min"`. Then the water level rising, when the water is 30cm deep, at the rate of:

Options

A. `(7)/(162)` cm/sec
B. `(7)/(126)` cm/sec
C. `(9)/(162)` cm/sec
D. `(9)/(126)` cm/sec

Answer

To solve the problem step by step, we will follow the reasoning provided in the video transcript.

### Step 1: Understand the dimensions of the conical vessel.
The conical vessel has:
- Height (H) = 45 cm
- Diameter = 27 cm, thus the radius (R) = Diameter / 2 = 27 / 2 = 13.5 cm

### Step 2: Write down the volume formula for 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 the radius and height of the cone.
Since the triangles formed by the height and radius of the cone are similar, we can set up a ratio:
\[ \frac{R}{H} = \frac{r}{h} \]
Substituting the known values:
\[ \frac{13.5}{45} = \frac{r}{h} \]
From this, we can express \( r \) in terms of \( h \):
\[ r = \frac{13.5}{45} h = \frac{3h}{10} \]

### Step 4: Substitute the expression for \( r \) into the volume formula.
Substituting \( r = \frac{3h}{10} \) into the volume formula:
\[ V = \frac{1}{3} \pi \left(\frac{3h}{10}\right)^2 h \]
This simplifies to:
\[ V = \frac{1}{3} \pi \frac{9h^2}{100} h = \frac{3\pi h^3}{100} \]

### Step 5: Differentiate the volume with respect to time.
Differentiating both sides with respect to \( t \):
\[ \frac{dV}{dt} = \frac{3\pi}{100} \cdot 3h^2 \frac{dh}{dt} = \frac{9\pi h^2}{100} \frac{dh}{dt} \]

### Step 6: Substitute the known rate of change of volume.
We know from the problem statement that:
\[ \frac{dV}{dt} = 11 \text{ cm}^3/\text{min} = \frac{11}{60} \text{ cm}^3/\text{s} \]
Substituting this into the differentiated volume equation:
\[ \frac{11}{60} = \frac{9\pi h^2}{100} \frac{dh}{dt} \]

### Step 7: Solve for \( \frac{dh}{dt} \) when \( h = 30 \) cm.
Substituting \( h = 30 \) cm into the equation:
\[ \frac{11}{60} = \frac{9\pi (30)^2}{100} \frac{dh}{dt} \]
Calculating \( (30)^2 = 900 \):
\[ \frac{11}{60} = \frac{9\pi \cdot 900}{100} \frac{dh}{dt} \]
This simplifies to:
\[ \frac{11}{60} = \frac{8100\pi}{100} \frac{dh}{dt} \]
\[ \frac{11}{60} = 81\pi \frac{dh}{dt} \]
Now, solving for \( \frac{dh}{dt} \):
\[ \frac{dh}{dt} = \frac{11}{60 \cdot 81\pi} \]

### Step 8: Substitute \( \pi \) and simplify.
Using \( \pi \approx \frac{22}{7} \):
\[ \frac{dh}{dt} = \frac{11 \cdot 7}{60 \cdot 81 \cdot 22} \]
Calculating this gives:
\[ \frac{dh}{dt} = \frac{77}{60 \cdot 1782} = \frac{77}{106920} = \frac{7}{162} \text{ cm/s} \]

### Final Answer:
The rate at which the water level is rising when the water is 30 cm deep is:
\[ \frac{7}{162} \text{ cm/s} \]