For closed organ pipe: Frequency of different modes, $\nu_{n}^{\prime}=\frac{(2 p-1) v}{4 L^{\prime}}$ where $(2 p-1)$ represents nth harmonic.
Thus frequency of nth harmonic in closed pipe, $\nu_{n}^{\prime}=\frac{n v}{4 L^{\prime}}$
Ist overtone i.e $p=2, \nu_{3}^{\prime}=\frac{3 v}{4 L^{\prime}}$
For open organ pipe: Frequency of mth harmonic $\nu_{m}=\frac{m v}{2 L}$
Ist overtone i.e $m=2, \nu_{2}=\frac{2 v}{2 L}$
Given $: \nu_{3}^{\prime}=\nu_{2}$
$\frac{3 v}{4 L^{\prime}}=\frac{v}{L} \Longrightarrow \frac{L^{\prime}}{L}=\frac{3}{4}…..(1)$
Also given, $\nu_{n}^{\prime}=\nu_{m}$
$\frac{n v}{4 L^{\prime}}=\frac{m v}{2 L}$
$\frac{n}{2}=m \frac{L^{\prime}}{L} \Longrightarrow \frac{n}{2}=m \frac{3}{4}$
Thus $2 n=3 m$
Now $n=9$ and $m=6$ satisfies the relation.
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Statement $-1$ : A capillary is dipped in a liquid and liquid rises to a height $h$ in it. As the temperature of the liquid is raised, the height $h$ increases (if the density of the liquid and the angle of contact remain the same).
Statement $-2$ : Surface tension of a liquid decreases with the rise in its temperature.