If water drops are falling at regular time intervals from ceiling of height H, then position of drop from the ceiling is (when `n^(th)` drop falling from the ceiling and `r^(th)` drop is in its way) 

Options

A. `((n-1)^2)/((n-r)^2) H`
B. `((n-r)^2)/((n-1)^2) H`
C. `((n-r)^2)/((n+1)^2) H`
D. `((n+1)^2)/((n+r)^2) H`

Answer

To solve the problem, we need to determine the position of the `r^(th)` drop from the ceiling when the `n^(th)` drop is falling. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the scenario
We have water drops falling from a ceiling of height \( H \) at regular time intervals. We need to find the position of the `r^(th)` drop from the ceiling when the `n^(th)` drop is falling.

### Step 2: Define the time intervals
Let’s denote:
- The time taken for the `n^(th)` drop to fall as \( t_n \).
- The time taken for the `r^(th)` drop to fall as \( t_r \).

Since the drops are falling at regular intervals, we can express:
- \( t_n = n \cdot \Delta t \) (where \( \Delta t \) is the time interval between drops)
- \( t_r = r \cdot \Delta t \)

### Step 3: Determine the distance fallen by the `n^(th)` drop
Using the equation of motion for free fall, the distance fallen by the `n^(th)` drop when it has been falling for \( t_n \) seconds is given by:
\[ h_n = \frac{1}{2} g t_n^2 = \frac{1}{2} g (n \cdot \Delta t)^2 \]

### Step 4: Determine the distance fallen by the `r^(th)` drop
Similarly, the distance fallen by the `r^(th)` drop when it has been falling for \( t_r \) seconds is:
\[ h_r = \frac{1}{2} g t_r^2 = \frac{1}{2} g (r \cdot \Delta t)^2 \]

### Step 5: Calculate the position of the `r^(th)` drop from the ceiling
The position of the `r^(th)` drop from the ceiling can be calculated as:
\[ \text{Position of } r^{th} \text{ drop} = H - h_r \]
Substituting \( h_r \):
\[ \text{Position of } r^{th} \text{ drop} = H - \frac{1}{2} g (r \cdot \Delta t)^2 \]

### Step 6: Relate \( \Delta t \) to the height \( H \)
We know that the first drop takes time \( t_1 = \sqrt{\frac{2H}{g}} \) to hit the ground. Since drops fall at regular intervals, we can express \( \Delta t \) in terms of \( H \):
\[ \Delta t = \frac{t_1}{n-1} = \frac{\sqrt{\frac{2H}{g}}}{n-1} \]

### Step 7: Substitute \( \Delta t \) back into the position equation
Now substituting \( \Delta t \) back into the position equation:
\[ \text{Position of } r^{th} \text{ drop} = H - \frac{1}{2} g \left(r \cdot \frac{\sqrt{\frac{2H}{g}}}{n-1}\right)^2 \]
This simplifies to:
\[ \text{Position of } r^{th} \text{ drop} = H - \frac{1}{2} g \cdot \frac{2Hr^2}{(n-1)^2} = H - \frac{gHr^2}{(n-1)^2} \]

### Final Result
Thus, the position of the `r^(th)` drop from the ceiling when the `n^(th)` drop is falling is:
\[ \text{Position of } r^{th} \text{ drop} = H - \frac{gHr^2}{(n-1)^2} \]