The first overtone of an open organ pipe ( of length $L_0$ ) beats with the first overtone of a closed organ pipe ( of length $L_c$) with a beat frequency of $10 Hz.$ The fundamental frequency of the closed pipe is $110 Hz$. If the speed of sound is $330 m{s}^{-1},$ then 

The fundamental frequencies of the open and closed pipes respectively are

$\begin{array}{l}{v}_0=\frac{\mathrm{v}}{2L_0}\\ AND v_{\mathrm{c}}=\frac{\mathrm{v}}{4L_c}\end{array}$

where v is the speed of sound. and In an open pipe, all harmonics are present. Hence the frequencies of the overtones are $2, 3, 4, .....$ etc. times the fundamental frequency. Hence the frequency of the first overtone in the open pipe is 

$v_1=2v_0=\frac{2\mathrm{v}}{2L_0}=\frac{\mathrm{v}}{L_0}$

In a closed pipe, only odd harmonics are present. Hence, the frequencies of the overtones are $3, 5, 7, ...$ etc., times the fundamental frequency. Hence the frequency of the first overtone in the closed pipe is

$v_2=3v_c=\frac{3\mathrm{v}}{4L_c}$

Now  $110=\frac{330}{4L_c}\Rightarrow L_{\mathrm{c}}=0.75\mathrm{m}.$

Given $v_1-v_2=\pm 10.$ There are the following two posibilities.

Case $\left(a\right):v_1-v_2=10.$ Thus,

$\frac{v}{L_0}-\frac{3v}{4L_c}=10$

Putting $\mathrm{v}=330{\mathrm{ms}}^{-1}\text{ and }{L}_{\mathrm{c}}=0.75\mathrm{m},$ we get

$L_0=\left(\frac{33}{34}\right)\mathrm{m}$

Case $\left(b\right):v_1-v_2=-10.$ In this case, we get

$L=\left(\frac{33}{32}\right)\mathrm{m}$

Hence the correct choices are (1) and (4).