A hollow metallic tube of length L and closed at one end produce resonance with a tuning fork of frequency n. The entire tube is then heated carefully so that at equilibrium temperature its length changes by `l`. If the change in velocity V of sound is v, the resonance will now produced by tuning fork of frequency :-

Options

A. `(V + v)//(4(L +l))`
B. `(V+v)//(4(L-l))`
C. `(V-v)//(4(L+l))`
D. `(V-v)//(4(L-l))`

Explanation

To solve the problem step by step, let's break it down as follows:

### Step 1: Understand the initial conditions
We have a hollow metallic tube of length \( L \) that is closed at one end. This tube produces resonance with a tuning fork of frequency \( n \).

### Step 2: Determine the wavelength in the initial condition
For a tube closed at one end, the fundamental frequency has a node at the closed end and an antinode at the open end. The relationship between the length of the tube and the wavelength \( \lambda \) is given by:
\[ \frac{\lambda}{4} = L \implies \lambda = 4L \]

### Step 3: Relate frequency to wavelength and velocity
The frequency \( f \) can be expressed as:
\[ f = \frac{v}{\lambda} \]
Substituting for \( \lambda \):
\[ f = \frac{v}{4L} \]
Since it is given that this frequency is equal to \( n \):
\[ n = \frac{v}{4L} \]

### Step 4: Analyze the changes after heating
When the tube is heated, its length changes by \( l \). Therefore, the new length of the tube becomes:
\[ L' = L + l \]

### Step 5: Determine the new wavelength
The new length of the tube still follows the same principle of resonance:
\[ \frac{\lambda'}{4} = L + l \implies \lambda' = 4(L + l) \]

### Step 6: Determine the new velocity of sound
The problem states that the change in the velocity of sound is \( v \), so the new velocity \( v' \) becomes:
\[ v' = v + v \]

### Step 7: Calculate the new frequency
Now we can find the new frequency \( f' \) using the new velocity and wavelength:
\[ f' = \frac{v'}{\lambda'} = \frac{v + v}{4(L + l)} = \frac{2v}{4(L + l)} = \frac{v}{2(L + l)} \]

### Step 8: Relate the new frequency to the original frequency
From the original frequency \( n = \frac{v}{4L} \), we can express \( v \) as:
\[ v = 4nL \]
Substituting this into the equation for \( f' \):
\[ f' = \frac{4nL}{2(L + l)} = \frac{2nL}{L + l} \]

### Conclusion
Thus, the frequency at which the tube will resonate after heating is:
\[ f' = \frac{2nL}{L + l} \]