Question

A closed organ pipe of length $100\text{cm}$ and an open organ pipe contain gases of densities $9{\text{kg/m}}^3$ and $1{\text{kg/m}}^3$, respectively. The compressibility of gases is equal in both the pipes. Both the pipes are vibrating in their first overtone with same frequency. The length of the open organ pipe is $\text{(in m)}$

• A. 4.0
• B. 4
• C. 4.00

Answer 1

Answer: (A), (B), (C)

Let $f_o$ be the frequency of first overtone of open organ pipe and $f_c$ be the frequency of first overtone for a closed organ pipe.
So $f_o=2\frac{v_o}{2L_o}$ and $f_c=3\frac{v_c}{4L_c}$
From the data given in the question, $f_o=f_c$

so, $\frac{2v_o}{2L_o}=\frac{3v_c}{4L_c}$
$\Rightarrow \frac{\sqrt{\frac{B_0}{{\rho }_o}}}{L_o}=\frac{3}{4}\frac{\sqrt{\frac{B_c}{{\rho }_c}}}{L_c}[\because v=\sqrt{\frac{B}{\rho }};B\text{is Bulk modulus of gas}]$

It is given that, the compressibility of two gases is same $\therefore B_0=B_c=B\text{(say)}$

$\Rightarrow \frac{1}{\sqrt{{\rho }_o}{L}_o}=\frac{3}{4}\frac{1}{\sqrt{{\rho }_c}{L}_c}$

Substituting the given data, we get
$\frac{1}{\sqrt{1}\times L_o}=\frac{3}{4}\times \frac{1}{\sqrt{9}\times 1}$

$\Rightarrow L_o=4\text{m}$