The oil is leaking from a drum at a rate of `16 cm^(3)//sec`. If the radius and height of drum are 7 cm and 60 cm respectively, find the rate of change of the height of oil when height of oil in drum is 18 cm.

Answer

To solve the problem, we need to find the rate of change of the height of oil in the drum when the height of the oil is 18 cm. We will use the formula for the volume of a cylinder and apply differentiation with respect to time.

### Step-by-step Solution:

1. **Identify the Variables**:
- Let \( V \) be the volume of the oil in the drum.
- The volume \( V \) of a cylinder is given by the formula:
\[ V = \pi r^2 h \]
- Here, \( r \) is the radius of the drum and \( h \) is the height of the oil.

2. **Given Values**:
- The radius \( r = 7 \) cm (constant, since the drum's radius does not change).
- The rate of change of volume \( \frac{dV}{dt} = -16 \) cm³/sec (negative because the volume is decreasing).
- The height of the oil \( h = 18 \) cm at the moment we want to find the rate of change of height.

3. **Differentiate the Volume with Respect to Time**:
- Differentiate \( V = \pi r^2 h \) with respect to \( t \):
\[ \frac{dV}{dt} = \pi r^2 \frac{dh}{dt} \]

4. **Substitute Known Values**:
- Substitute \( r = 7 \) cm into the differentiated equation:
\[ \frac{dV}{dt} = \pi (7^2) \frac{dh}{dt} \]
- This simplifies to:
\[ \frac{dV}{dt} = \pi (49) \frac{dh}{dt} \]
- Therefore:
\[ \frac{dV}{dt} = 49\pi \frac{dh}{dt} \]

5. **Set Up the Equation**:
- Now, we know \( \frac{dV}{dt} = -16 \):
\[ -16 = 49\pi \frac{dh}{dt} \]

6. **Solve for \( \frac{dh}{dt} \)**:
- Rearranging gives:
\[ \frac{dh}{dt} = \frac{-16}{49\pi} \]

7. **Calculate \( \frac{dh}{dt} \)**:
- Using \( \pi \approx \frac{22}{7} \):
\[ \frac{dh}{dt} = \frac{-16}{49 \times \frac{22}{7}} = \frac{-16 \times 7}{49 \times 22} = \frac{-112}{1078} \]
- Simplifying gives:
\[ \frac{dh}{dt} \approx -\frac{56}{539} \text{ cm/sec} \]

### Final Answer:
The rate of change of the height of oil when the height is 18 cm is approximately:
\[ \frac{dh}{dt} \approx -\frac{56}{539} \text{ cm/sec} \]