A solid cylinder of height H has a conical portion of same height and radius `1//3rd` of height removed from it.
Rain water is accumulating in it, at the rate equal to `pi` times the instaneous radius of the water surface inside the hole, the time after which hole will filled with water is

Options

A. `(H^(2))/(3)`
B. `H^(2)`
C. `(H^(2))/(6)`
D. `H^(2)/(4)`

Answer

To solve the problem, we need to find the time after which the conical portion of the cylinder will be filled with rainwater. Let's break down the solution step by step.

### Step 1: Understand the Geometry
We have a solid cylinder of height \( H \) and a conical portion removed from it. The radius of the cone is \( \frac{H}{3} \) (one-third of the height).

### Step 2: Volume of the Conical Portion
The volume \( V \) of a cone is given by the formula:
\[ V = \frac{1}{3} \pi r^2 h \]
Here, the radius \( r \) of the cone is \( \frac{H}{3} \) and the height \( h \) is \( H \). Substituting these values:
\[ V = \frac{1}{3} \pi \left(\frac{H}{3}\right)^2 H = \frac{1}{3} \pi \left(\frac{H^2}{9}\right) H = \frac{1}{27} \pi H^3 \]

### Step 3: Rate of Accumulation of Water
The rate of accumulation of rainwater is given as \( \pi \) times the instantaneous radius of the water surface inside the hole. Let the instantaneous radius of the water surface be \( r \). Therefore, the rate of volume increase is:
\[ \frac{dV}{dt} = \pi r \]

### Step 4: Relate Radius and Height
Since the cone's dimensions are proportional, we can express the radius \( r \) in terms of the height \( h \) of the water:
\[ \frac{r}{h} = \frac{H/3}{H} \implies r = \frac{h}{3} \]

### Step 5: Substitute \( r \) in the Volume Rate
Substituting \( r = \frac{h}{3} \) into the rate of volume increase:
\[ \frac{dV}{dt} = \pi \left(\frac{h}{3}\right) = \frac{\pi h}{3} \]

### Step 6: Volume of Water in the Cone
The volume of water in the cone when the height is \( h \) is:
\[ V = \frac{1}{3} \pi r^2 h = \frac{1}{3} \pi \left(\frac{h}{3}\right)^2 h = \frac{1}{3} \pi \left(\frac{h^2}{9}\right) h = \frac{1}{27} \pi h^3 \]

### Step 7: Differentiate Volume with Respect to Time
Differentiating \( V \) with respect to time:
\[ \frac{dV}{dt} = \frac{1}{27} \pi (3h^2 \frac{dh}{dt}) = \frac{\pi h^2}{9} \frac{dh}{dt} \]

### Step 8: Set the Two Expressions for \( \frac{dV}{dt} \) Equal
Now, we can set the two expressions for \( \frac{dV}{dt} \) equal to each other:
\[ \frac{\pi h^2}{9} \frac{dh}{dt} = \frac{\pi h}{3} \]
Canceling \( \pi \) from both sides:
\[ \frac{h^2}{9} \frac{dh}{dt} = \frac{h}{3} \]
Multiplying both sides by 9:
\[ h^2 \frac{dh}{dt} = 3h \]
Dividing both sides by \( h \) (assuming \( h \neq 0 \)):
\[ h \frac{dh}{dt} = 3 \]

### Step 9: Separate Variables and Integrate
Separating variables:
\[ h \, dh = 3 \, dt \]
Integrating both sides:
\[ \int h \, dh = \int 3 \, dt \]
This gives:
\[ \frac{h^2}{2} = 3t + C \]
At \( t = 0 \), \( h = 0 \) implies \( C = 0 \):
\[ \frac{h^2}{2} = 3t \implies h^2 = 6t \]

### Step 10: Find Time When Cone is Full
When the cone is full, \( h = H \):
\[ H^2 = 6t \implies t = \frac{H^2}{6} \]

### Final Answer
The time after which the hole will be filled with water is:
\[ t = \frac{H^2}{6} \]