Two identical strings X and Z made of same material have tension $$T_X$$ and $$T_Z$$ in them. If their fundamental frequencies are 450 Hz and 300 Hz, respectively, then the ratio $$T_X/T_Z$$ is:

We are told that both strings X and Z are identical, which means they have the same length $$L$$, the same material, and therefore the same linear mass density $$\mu$$. For a stretched string the fundamental (first-harmonic) frequency is given by the standard formula

$$f \;=\;\dfrac{1}{2L}\;\sqrt{\dfrac{T}{\mu}}$$

Here $$f$$ is the fundamental frequency, $$T$$ is the tension in the string and $$\mu$$ is its mass per unit length. Because the strings are identical, the quantities $$L$$ and $$\mu$$ are the same for both strings. Let us write the formula separately for strings X and Z.

For string X we have

$$f_X \;=\;\dfrac{1}{2L}\;\sqrt{\dfrac{T_X}{\mu}}$$

For string Z we have

$$f_Z \;=\;\dfrac{1}{2L}\;\sqrt{\dfrac{T_Z}{\mu}}$$

Now we take the ratio of the two frequencies. Dividing the first equation by the second, the common factor $$1/(2L)$$ cancels out immediately:

$$\dfrac{f_X}{f_Z}\;=\;\dfrac{\sqrt{T_X/\mu}}{\sqrt{T_Z/\mu}} =\sqrt{\dfrac{T_X}{T_Z}}$$

We want the ratio $$T_X/T_Z$$, so we square both sides of the last equation:

$$(\dfrac{f_X}{f_Z})^{2} \;=\;\dfrac{T_X}{T_Z}$$

Now we substitute the given numerical values. The fundamental frequency of X is 450 Hz and that of Z is 300 Hz, hence

$$\dfrac{f_X}{f_Z}\;=\;\dfrac{450}{300}\;=\;1.5$$

Substituting this ratio into the squared relation, we get

$$\dfrac{T_X}{T_Z}\;=\;(1.5)^{2}$$

Evaluating the square,

$$\dfrac{T_X}{T_Z}\;=\;2.25$$

Hence, the correct answer is Option A.