Water pours out rate of Q from a tap, into a cylindrical vessel of radius r. The rate at which the height of water level rises the height is h, is

Options

A. `Q/(pirh)`
B. `Q/(pir^(2))`
C. `Q/(2pir^(2))`
D. `Q/(pir^(2)h)`

Explanation

To solve the problem, we need to determine the rate at which the height of water in a cylindrical vessel rises when water is poured into it at a rate \( Q \).

### Step-by-Step Solution:

1. **Identify the Variables:**
- Let \( Q \) be the rate at which water is flowing out of the tap (in cubic meters per second).
- Let \( r \) be the radius of the cylindrical vessel.
- Let \( h \) be the height of the water in the cylindrical vessel.
- We need to find the rate of change of height, \( \frac{dh}{dt} \).

2. **Volume of Water in the Cylinder:**
- The volume \( V \) of water in a cylindrical vessel is given by the formula:
\[ V = \pi r^2 h \]
- Here, \( \pi r^2 \) is the cross-sectional area of the cylinder, and \( h \) is the height of the water.

3. **Rate of Change of Volume:**
- The rate at which the volume of water in the cylinder is changing with respect to time is equal to the rate at which water is being poured in, which is \( Q \):
\[ \frac{dV}{dt} = Q \]

4. **Differentiate the Volume with Respect to Time:**
- To find the relationship between the rate of change of height and the rate of change of volume, we differentiate the volume equation with respect to time \( t \):
\[ \frac{dV}{dt} = \frac{d}{dt}(\pi r^2 h) = \pi r^2 \frac{dh}{dt} \]
- Here, \( \pi r^2 \) is a constant since the radius \( r \) does not change.

5. **Set the Two Rates Equal:**
- Since both expressions represent the rate of change of volume, we can set them equal to each other:
\[ Q = \pi r^2 \frac{dh}{dt} \]

6. **Solve for \( \frac{dh}{dt} \):**
- Rearranging the equation to solve for \( \frac{dh}{dt} \):
\[ \frac{dh}{dt} = \frac{Q}{\pi r^2} \]

### Final Answer:
The rate at which the height of the water level rises is given by:
\[ \frac{dh}{dt} = \frac{Q}{\pi r^2} \]