Two closed organ pipes $A$ and $B$ have the same length. $A$ is wider than $B$. They resonate in the…
Two closed organ pipes $A$ and $B$ have the same length. $A$ is wider than $B$. They resonate in the fundamental mode at frequencies $n_A$ and $n_B$ respectively, then
$n_A = n_B$
$n_A > n_B$
$n_A < n_B$
Either (b) or (c) depending on the ratio of their diameter
Solution
In closed organ pipe, first resonance occurs at $\lambda / 4$. So, in fundamental mode of vibration of organ pipe $\lambda / 4 = (l + 0.3d)$, where $0.3 d$ is necessary end correction, then frequency of vibration, $n = \frac{v}{\lambda} = \frac{v}{4(l + 0.3d)}$.
As $l$ is same, wide pipe $A$ will resonate at a lower frequency.
So, $n_A < n_B$.