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Q.
The end correction of a resonance column is $1 \,cm$. If the shortest length resonating with the tuning fork is $10\, cm$, the next resonating length should be :
Let $x$ be the end correction and $l_{2}$ be the second resonance length.
Given: first resonance length, $l_{1}=10 cm$ and end correction of a resonance column, $x=1 cm$.
For first resonance, we have
$\frac{\lambda}{4}=l_{1}+x$
$\frac{\lambda}{4}=10+1=11 cm$
$\lambda=44 cm$
For next resonance, that is, second resonance,
$\frac{3 \lambda}{4} =l_{2}+x$
$3 \times 11 =l_{2}+1$ or $33=l_{2}+1$
$l_{2} =32\, cm$
