A cylindrical tank of height 50 cm is completely filled with water. On it’s vertical side two small holes are there, such that water flowing out of both the holes strikes the ground at same point. If first hole is at depth 30 cm from the top, then the depth of second hole from top will be :

Options

A. 10 cm
B. 20 cm
C. 25 cm
D. 40 cm

Answer

To solve the problem, we need to determine the depth of the second hole from the top of the cylindrical tank, given that both holes allow water to flow out and strike the ground at the same point.

### Step-by-Step Solution:

1. **Understand the Setup**:
- The cylindrical tank has a height of 50 cm and is filled with water.
- The first hole is located 30 cm from the top, which means it is 20 cm from the bottom (50 cm - 30 cm = 20 cm).

2. **Identify the Depth of the Second Hole**:
- Let the depth of the second hole from the top be \( h \) cm. Consequently, its depth from the bottom will be \( 50 - h \) cm.

3. **Use the Range Formula**:
- The range \( R \) of water flowing out of a hole in a tank can be expressed as:
\[ R = 2 \sqrt{h_1 \cdot h_2} \]
- For the first hole (at 20 cm from the bottom):
\[ R_1 = 2 \sqrt{20 \cdot 30} \]
- For the second hole (at \( 50 - h \) cm from the bottom):
\[ R_2 = 2 \sqrt{(50 - h) \cdot h} \]

4. **Set the Ranges Equal**:
- Since both holes strike the ground at the same point, we can set the ranges equal to each other:
\[ 2 \sqrt{20 \cdot 30} = 2 \sqrt{(50 - h) \cdot h} \]
- We can simplify this equation by dividing both sides by 2:
\[ \sqrt{20 \cdot 30} = \sqrt{(50 - h) \cdot h} \]

5. **Square Both Sides**:
- Squaring both sides to eliminate the square root gives:
\[ 20 \cdot 30 = (50 - h) \cdot h \]
- This simplifies to:
\[ 600 = 50h - h^2 \]

6. **Rearranging the Equation**:
- Rearranging gives us a quadratic equation:
\[ h^2 - 50h + 600 = 0 \]

7. **Using the Quadratic Formula**:
- We can solve for \( h \) using the quadratic formula:
\[ h = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
- Here, \( a = 1 \), \( b = -50 \), and \( c = 600 \):
\[ h = \frac{50 \pm \sqrt{(-50)^2 - 4 \cdot 1 \cdot 600}}{2 \cdot 1} \]
\[ h = \frac{50 \pm \sqrt{2500 - 2400}}{2} \]
\[ h = \frac{50 \pm \sqrt{100}}{2} \]
\[ h = \frac{50 \pm 10}{2} \]

8. **Calculating the Values**:
- This gives us two possible values for \( h \):
\[ h = \frac{60}{2} = 30 \quad \text{and} \quad h = \frac{40}{2} = 20 \]

9. **Selecting the Correct Depth**:
- Since the first hole is at 30 cm from the top, the second hole must be at 20 cm from the top.

### Final Answer:
The depth of the second hole from the top is **20 cm**.