A string is stretched between two fixed points separated by $75.0 \mathrm{cm}.$ It is observed to have resonant frequencies of $420 \mathrm{Hz}$ and $315 \mathrm{Hz}.$There are no other resonant frequencies between these two. The lowest resonant frequency for this string is:

In a stretched string between two fixed ends, all multiples of frequencies can be obtained i.e., if fundamental frequency is n then higher frequencies will be $2\mathrm{n}, 3\mathrm{n}, 4\mathrm{n} ...$

So, the difference between any two successive frequencies will be $\text{'}\mathrm{n}\text{'}$

According to question, $\mathrm{n} = 420 - 315 = 105 \mathrm{Hz}$

So the lowest frequency of the string is $105 \mathrm{Hz}.$