Question

A hollow cylinder with both sides open generates a frequency $ f $ in air. When the cylinder vertically immersed into water by half its length the frequency will be

• A. f
• B. 2f
• C. f/2
• D. f/4

Answer 1

The Correct Option is A

(i) Here, $\frac{\lambda}{2}=l$
$\Rightarrow \lambda=2 l$
So, $v=\frac{v}{21}$

(ii) and $\frac{\lambda}{4}=\frac{l}{2}$
$\lambda=\frac{4 l}{2}=2 l$
$\therefore v_{2}=\frac{v}{2 l}$, the same.
$\therefore f$ (original) is the frequency.

Answer 2

Let's analyze the problem step by step.
 Given:
- A hollow cylinder with both sides open generates a frequency \( f \) in air.
- The cylinder is vertically immersed into water by half its length.
- We need to determine the new frequency.
Solution:
1. Initial Frequency (in Air):
  - The cylinder is open at both ends, which means it acts as an open pipe.
  - For an open pipe, the fundamental frequency (\( f \)) is given by:
    \[ f = \frac{v}{2L}  \]
    where \( v \) is the speed of sound in air and \( L \) is the length of the cylinder.
2. Immersed Cylinder (Half in Water):
  - When the cylinder is immersed in water by half its length, one end remains in the air, and the other end is in water.
  - The length of the air column inside the cylinder is now \( \frac{L}{2} \).
3. Frequency in the New Situation:
  - The cylinder now has a different boundary condition: one end is closed by water (acting as a closed end), and the other end is open to the air.
  - For a pipe with one end closed and the other end open, the fundamental frequency is given by:
    \[ f' = \frac{v}{4L'}\]
    where \( L' \) is the length of the air column, which is \( \frac{L}{2} \).
4. Calculating the New Frequency:
  \[  f' = \frac{v}{4 \left(\frac{L}{2}\right)} = \frac{v}{2L} = f \]
  Thus, the frequency remains \( f \).
Conclusion:
When the cylinder is vertically immersed in water by half its length, the new frequency generated is the same as the original frequency \( f \).
Therefore, the correct answer is option (A): \( f \).