Water is dropped at the rate of `2m^(2)//s` into a cone of semivertical angel of `45^(@)`. The rate at which periphery of water surface changes when height of water in the cone is 2 m, is

Options

A. `0.5m//s`
B. `2m//s`
C. `3m//s`
D. `1m//s`

Explanation

To solve the problem step by step, we will follow the reasoning laid out in the video transcript.

### Step 1: Understand the Geometry of the Cone
We have a cone with a semi-vertical angle of \(45^\circ\). This means that the radius \(r\) of the water surface and the height \(h\) of the water are equal when the angle is \(45^\circ\). Therefore, we can express the radius in terms of height:
\[ r = h \]

### Step 2: Volume of the Cone
The volume \(V\) of a cone is given by the formula:
\[ V = \frac{1}{3} \pi r^2 h \]
Substituting \(r = h\) into the volume formula:
\[ V = \frac{1}{3} \pi h^2 h = \frac{1}{3} \pi h^3 \]

### Step 3: Differentiate the Volume with Respect to Time
To find the rate at which the volume changes with respect to time, we differentiate \(V\) with respect to \(t\):
\[ \frac{dV}{dt} = \frac{d}{dt} \left(\frac{1}{3} \pi h^3\right) = \pi h^2 \frac{dh}{dt} \]

### Step 4: Relate the Rate of Volume Change to the Given Rate
We know that water is being added at a rate of \(2 \, m^3/s\), so:
\[ \frac{dV}{dt} = 2 \]
Setting this equal to our differentiated volume:
\[ 2 = \pi h^2 \frac{dh}{dt} \]

### Step 5: Solve for \(\frac{dh}{dt}\)
We need to find \(\frac{dh}{dt}\) when \(h = 2 \, m\):
\[ 2 = \pi (2^2) \frac{dh}{dt} \]
\[ 2 = 4\pi \frac{dh}{dt} \]
\[ \frac{dh}{dt} = \frac{2}{4\pi} = \frac{1}{2\pi} \, m/s \]

### Step 6: Find the Rate of Change of the Radius
Since \(r = h\), we have:
\[ \frac{dr}{dt} = \frac{dh}{dt} = \frac{1}{2\pi} \, m/s \]

### Step 7: Find the Rate of Change of the Periphery
The perimeter \(P\) of the water surface is given by:
\[ P = 2\pi r \]
Differentiating with respect to time:
\[ \frac{dP}{dt} = 2\pi \frac{dr}{dt} \]
Substituting \(\frac{dr}{dt}\):
\[ \frac{dP}{dt} = 2\pi \left(\frac{1}{2\pi}\right) = 1 \, m/s \]

### Final Answer
The rate at which the periphery of the water surface changes when the height of the water in the cone is 2 m is:
\[ \frac{dP}{dt} = 1 \, m/s \]