Water is dropped at the rate of 2 `m^3`/s into a cone of semi-vertical angle is `45^@` . If the rate at which periphery of water surface changes when the height of the water in the cone is 2m is d. Then the value of 5d is _____ m/sec

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the Geometry of the Cone
Given that the cone has a semi-vertical angle of 45 degrees, we can establish a relationship between the radius (r) and height (h) of the water in the cone.

Since the angle is 45 degrees, we have:
\[ \tan(45^\circ) = 1 = \frac{r}{h} \implies r = h \]

### Step 2: Write the 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, we get:
\[ V = \frac{1}{3} \pi h^2 h = \frac{1}{3} \pi h^3 \]

### Step 3: Differentiate the Volume with Respect to Time
We know that water is being added at a rate of \( \frac{dV}{dt} = 2 \, m^3/s \). Now, we differentiate the volume with respect to time \( t \):
\[ \frac{dV}{dt} = \frac{d}{dt}\left(\frac{1}{3} \pi h^3\right) = \pi h^2 \frac{dh}{dt} \]
Setting this equal to the rate of volume change:
\[ \pi h^2 \frac{dh}{dt} = 2 \]

### Step 4: Solve for \( \frac{dh}{dt} \)
Now we can solve for \( \frac{dh}{dt} \):
\[ \frac{dh}{dt} = \frac{2}{\pi h^2} \]

### Step 5: Find the Radius at the Given Height
At the height \( h = 2 \, m \):
\[ r = h = 2 \, m \]

### Step 6: Calculate \( \frac{dh}{dt} \) at \( h = 2 \)
Substituting \( h = 2 \) into the equation for \( \frac{dh}{dt} \):
\[ \frac{dh}{dt} = \frac{2}{\pi (2)^2} = \frac{2}{\pi \cdot 4} = \frac{1}{2\pi} \]

### Step 7: Find the Rate of Change of the Radius
Since \( r = h \), we have:
\[ \frac{dr}{dt} = \frac{dh}{dt} = \frac{1}{2\pi} \]

### Step 8: Calculate the Rate of Change of the Perimeter
The perimeter \( P \) of the water surface (which is a circle) is given by:
\[ P = 2\pi r \]
Differentiating with respect to time:
\[ \frac{dP}{dt} = 2\pi \frac{dr}{dt} \]
Substituting \( \frac{dr}{dt} = \frac{1}{2\pi} \):
\[ \frac{dP}{dt} = 2\pi \cdot \frac{1}{2\pi} = 1 \, m/s \]

### Step 9: Find the Value of \( 5d \)
Since we have \( \frac{dP}{dt} = d = 1 \, m/s \), we need to find \( 5d \):
\[ 5d = 5 \cdot 1 = 5 \, m/s \]

### Final Answer
Thus, the value of \( 5d \) is:
\[ \boxed{5} \, m/s \]