From a cylindrical drum containing oil and kept vertical, the oil leaking at the rate of `10cm^(3)//s.` If the radius of the durm is 10 cm and height is 50 cm, then find the rate at which level of oil is changing when oil level is 20cm.

Answer

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

### Step 1: Understand the problem
We have a cylindrical drum with a radius of \( r = 10 \, \text{cm} \) and a height of \( h = 50 \, \text{cm} \). Oil is leaking from the drum at a rate of \( \frac{dV}{dt} = -10 \, \text{cm}^3/\text{s} \) (negative because the volume is decreasing). We need to find the rate at which the height of the oil is changing, \( \frac{dh}{dt} \), when the height of the oil is \( h = 20 \, \text{cm} \).

### Step 2: Write the formula for the volume of a cylinder
The volume \( V \) of a cylinder is given by the formula:
\[ V = \pi r^2 h \]
where \( r \) is the radius and \( h \) is the height of the oil.

### Step 3: Differentiate the volume with respect to time
Since both the volume and height change with time, we differentiate the volume with respect to time \( t \):
\[ \frac{dV}{dt} = \pi r^2 \frac{dh}{dt} \]
Here, \( r \) is constant because the radius of the drum does not change.

### Step 4: Substitute known values
We know:
- \( \frac{dV}{dt} = -10 \, \text{cm}^3/\text{s} \)
- \( r = 10 \, \text{cm} \)

Substituting these values into the differentiated volume equation:
\[ -10 = \pi (10^2) \frac{dh}{dt} \]
This simplifies to:
\[ -10 = 100\pi \frac{dh}{dt} \]

### Step 5: Solve for \( \frac{dh}{dt} \)
Now, we can solve for \( \frac{dh}{dt} \):
\[ \frac{dh}{dt} = \frac{-10}{100\pi} = \frac{-1}{10\pi} \, \text{cm/s} \]

### Conclusion
The rate at which the level of oil is changing when the oil level is \( 20 \, \text{cm} \) is:
\[ \frac{dh}{dt} = -\frac{1}{10\pi} \, \text{cm/s} \]