The tension of a stretched string is increased by $69\%$. In order to keep its frequency of vibration constant,its length must be increased by .... $\%$

$A$ uniform string of length $L$ and mass $M$ is fixed at both ends while it is subject to a tension $T$. It can vibrate at frequencies $(v)$ given by the formula (where $n=1, 2, 3, \ldots$):

$A$ string is stretched between two rigid supports separated by $75 \,cm$. There are no resonant frequencies between $420 \,Hz$ and $315 \,Hz$. The lowest resonant frequency for the string is (in $\,Hz$)

$A$ stretched wire of length $110 \, cm$ is divided into three segments whose frequencies are in the ratio $1 : 2 : 3$. Their lengths must be:

$A$ string fixed at both ends vibrates in three loops. The wavelength is $10 \, cm$. The length of the string is .... $cm$.

$A$ sonometer wire of resonating length $90 \ cm$ has a fundamental frequency of $400 \ Hz$ when kept under some tension. The resonating length of the wire with a fundamental frequency of $600 \ Hz$ under the same tension is . . . . . . $cm$.