The rate at which fluid level inside vertical cylindrical tank of radius r drop if we pump fluid out at the rate of `3cm^(3)` /min is

Options

A. `-(1)/(pi r^(2))`
B. `(3)/(pi r^(2))`
C. `(2)/(pi r^(2))`
D. `(4)/(pi r)`

Answer

To find the rate at which the fluid level inside a vertical cylindrical tank drops when fluid is pumped out at a rate of 3 cm³/min, we can follow these steps:

### Step 1: Understand the Volume of the Cylinder
The volume \( V \) of a cylinder is given by the formula:
\[ V = \pi r^2 h \]
where \( r \) is the radius of the base of the cylinder and \( h \) is the height of the fluid in the cylinder.

### Step 2: Differentiate the Volume with Respect to Time
Since the radius \( r \) is constant and we are interested in how the height \( h \) changes over time as fluid is pumped out, we differentiate the volume with respect to time \( t \):
\[ \frac{dV}{dt} = \pi r^2 \frac{dh}{dt} \]

### Step 3: Set Up the Equation with Given Rate
We know that fluid is being pumped out at a rate of \( \frac{dV}{dt} = -3 \) cm³/min (the negative sign indicates a decrease in volume). Thus, we can write:
\[ -3 = \pi r^2 \frac{dh}{dt} \]

### Step 4: Solve for \( \frac{dh}{dt} \)
Rearranging the equation to solve for \( \frac{dh}{dt} \):
\[ \frac{dh}{dt} = \frac{-3}{\pi r^2} \]

### Conclusion
The rate at which the fluid level inside the vertical cylindrical tank drops is:
\[ \frac{dh}{dt} = \frac{-3}{\pi r^2} \text{ cm/min} \]