A cone whose height is always equal to its diameter is increasing in volume at the rate of `40 cm^(3) ` /sec . The rate at which its radius is increasing when its circular base area is ` 1 m^(2)` is

Options

A. 1mm/sec
B. 2mm/sec
C. 0.001 cm/sec
D. 0.002 cm/sec

Answer

To solve the problem, we need to find the rate at which the radius of a cone is increasing when its base area is 1 m², given that the volume of the cone is increasing at a rate of 40 cm³/sec and that the height of the cone is equal to its diameter.

### Step-by-Step Solution:

1. **Understand the relationship between height and radius**:
Given that the height \( h \) is equal to the diameter, we can express the height in terms of the radius:
\[ h = 2r \]

2. **Volume of the cone**:
The formula for the volume \( V \) of a cone is:
\[ V = \frac{1}{3} \pi r^2 h \]
Substituting \( h = 2r \):
\[ V = \frac{1}{3} \pi r^2 (2r) = \frac{2}{3} \pi r^3 \]

3. **Differentiate the volume with respect to time**:
To find how the volume changes over time, we differentiate \( V \) with respect to \( t \):
\[ \frac{dV}{dt} = \frac{d}{dt} \left( \frac{2}{3} \pi r^3 \right) = \frac{2}{3} \pi \cdot 3r^2 \frac{dr}{dt} = 2\pi r^2 \frac{dr}{dt} \]

4. **Set the rate of volume change**:
We know the rate of volume change \( \frac{dV}{dt} = 40 \) cm³/sec. Therefore, we can set up the equation:
\[ 40 = 2\pi r^2 \frac{dr}{dt} \]

5. **Convert the area of the base**:
The area of the base \( A \) is given as 1 m². Since the area of the base of the cone is \( \pi r^2 \):
\[ \pi r^2 = 1 \quad \Rightarrow \quad r^2 = \frac{1}{\pi} \]
Thus, we can find \( r \):
\[ r = \sqrt{\frac{1}{\pi}} \]

6. **Substituting \( r \) into the volume change equation**:
Now substitute \( r \) back into the equation:
\[ 40 = 2\pi \left(\frac{1}{\pi}\right) \frac{dr}{dt} \]
Simplifying this gives:
\[ 40 = 2 \frac{dr}{dt} \]

7. **Solve for \( \frac{dr}{dt} \)**:
Rearranging the equation:
\[ \frac{dr}{dt} = \frac{40}{2} = 20 \text{ cm/sec} \]

### Final Answer:
The rate at which the radius is increasing when the circular base area is 1 m² is:
\[ \frac{dr}{dt} = 20 \text{ cm/sec} \]