Water is poured into an inverted conical vessel of which the radius of the base is 2m and height 4m, at the rate of 77 lit/min. The rate at which the water level is rising, at the instant when the depth is 70 cm is

Options

A. (a) `10 cm//min`
B. (b) `20 cm//min`
C. (c) `40 cm//min`
D. (d) `30 cm//min`

Answer

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

### Step 1: Understand the problem
We have an inverted conical vessel with a radius of 2 meters and a height of 4 meters. Water is being poured into this vessel at a rate of 77 liters per minute. We need to find the rate at which the water level is rising when the depth of the water is 70 cm.

### Step 2: Convert units
Since the depth is given in centimeters, we will convert the rate of water being poured into the vessel from liters to cubic centimeters.

1 liter = 1000 cubic centimeters, so:
\[ \frac{dV}{dt} = 77 \text{ liters/min} = 77 \times 1000 \text{ cm}^3/\text{min} = 77000 \text{ cm}^3/\text{min} \]

### Step 3: Relate the radius and height of the cone
Using similar triangles, we can relate the radius \( r \) and height \( h \) of the water in the cone:
\[ \frac{r}{h} = \frac{2}{4} \implies r = \frac{h}{2} \]

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

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

### Step 6: Substitute known values
We know \( \frac{dV}{dt} = 77000 \text{ cm}^3/\text{min} \) and we want to find \( \frac{dh}{dt} \) when \( h = 70 \text{ cm} \):
\[ 77000 = \frac{\pi}{4} (70)^2 \frac{dh}{dt} \]

### Step 7: Solve for \( \frac{dh}{dt} \)
First, calculate \( (70)^2 \):
\[ (70)^2 = 4900 \]
Now substitute this into the equation:
\[ 77000 = \frac{\pi}{4} \cdot 4900 \cdot \frac{dh}{dt} \]
Rearranging gives:
\[ \frac{dh}{dt} = \frac{77000 \cdot 4}{\pi \cdot 4900} \]
Calculating the right side:
\[ \frac{dh}{dt} = \frac{308000}{\pi \cdot 4900} \]
Using \( \pi \approx \frac{22}{7} \):
\[ \frac{dh}{dt} = \frac{308000 \cdot 7}{22 \cdot 4900} \]

### Step 8: Simplify and calculate
Calculating the denominator:
\[ 22 \cdot 4900 = 107800 \]
Thus:
\[ \frac{dh}{dt} = \frac{308000 \cdot 7}{107800} \]
Now, simplifying:
\[ \frac{dh}{dt} = \frac{2156000}{107800} \approx 20 \text{ cm/min} \]

### Conclusion
The rate at which the water level is rising when the depth is 70 cm is approximately **20 cm/min**.