Points ` P, Q` and `R` are in vertical line such that ` PQ =QR`. A ball at (P) is allowed to fall freely. What is the ratio of the times of descent through PQ and QR ?

Options

A. 0.042361111111111
B. 1 : `sqrt 2`
C. `1 : (sqrt 2` - 1)
D. 1 : (`sqrt 2` + 1)

Explanation

To solve the problem, we need to find the ratio of the times of descent through the segments PQ and QR, given that the distances PQ and QR are equal. Let's denote the distance PQ = QR = H.

### Step-by-Step Solution:

1. **Define the Variables:**
- Let the distance between points P and Q be \( H \).
- Let the distance between points Q and R also be \( H \).
- Let \( t_1 \) be the time taken to fall from P to Q.
- Let \( t_2 \) be the time taken to fall from Q to R.

2. **Use the Equation of Motion for PQ:**
- Since the ball is dropped from rest, the initial velocity \( u = 0 \).
- The equation of motion for the distance \( H \) can be written as:
\[ H = ut_1 + \frac{1}{2} g t_1^2 \]
- Substituting \( u = 0 \):
\[ H = \frac{1}{2} g t_1^2 \]

3. **Use the Equation of Motion for PR:**
- The total distance from P to R is \( 2H \).
- The time taken to fall from P to R is \( t_1 + t_2 \).
- The equation of motion for the distance \( 2H \) can be written as:
\[ 2H = u(t_1 + t_2) + \frac{1}{2} g (t_1 + t_2)^2 \]
- Again substituting \( u = 0 \):
\[ 2H = \frac{1}{2} g (t_1 + t_2)^2 \]

4. **Relate the Two Equations:**
- From the first equation, we can express \( H \):
\[ H = \frac{1}{2} g t_1^2 \implies 2H = g t_1^2 \]
- Substitute \( 2H \) in the second equation:
\[ g t_1^2 = \frac{1}{2} g (t_1 + t_2)^2 \]
- Dividing both sides by \( g \) (assuming \( g \neq 0 \)):
\[ t_1^2 = \frac{1}{2} (t_1 + t_2)^2 \]

5. **Simplify the Equation:**
- Multiply both sides by 2:
\[ 2t_1^2 = (t_1 + t_2)^2 \]
- Expanding the right side:
\[ 2t_1^2 = t_1^2 + 2t_1t_2 + t_2^2 \]
- Rearranging gives:
\[ t_1^2 - 2t_1t_2 - t_2^2 = 0 \]

6. **Solve the Quadratic Equation:**
- This is a quadratic equation in terms of \( t_1 \):
\[ t_1^2 - 2t_1t_2 - t_2^2 = 0 \]
- Using the quadratic formula \( t_1 = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
- Here, \( a = 1, b = -2t_2, c = -t_2^2 \):
\[ t_1 = \frac{2t_2 \pm \sqrt{(2t_2)^2 - 4(-t_2^2)}}{2} \]
\[ t_1 = t_2 \pm t_2\sqrt{2} \]

7. **Find the Ratio \( \frac{t_1}{t_2} \):**
- We can express \( t_1 \) in terms of \( t_2 \):
\[ t_1 = t_2(\sqrt{2} - 1) \]
- Thus, the ratio \( \frac{t_1}{t_2} \) is:
\[ \frac{t_1}{t_2} = \sqrt{2} - 1 \]

### Final Answer:
The ratio of the times of descent through PQ and QR is:
\[ \frac{t_1}{t_2} = \sqrt{2} - 1 \]