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 fundamental frequency of $$600$$ Hz under same tension ______ cm.

We are given that a sonometer wire of resonating length 90 cm has a fundamental frequency of 400 Hz under a certain tension. We need to find the resonating length when the fundamental frequency is 600 Hz under the same tension.

Recall that the fundamental frequency of a string fixed at both ends is given by:

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

Here $$L$$ is the vibrating length, $$T$$ is the tension, and $$\mu$$ is the linear mass density of the wire. Since both $$T$$ and $$\mu$$ remain constant, it follows that $$f \propto \frac{1}{L}$$, which means $$f \times L$$ is constant, or

$$f_1 L_1 = f_2 L_2$$.

Substituting the given values $$f_1 = 400$$ Hz, $$L_1 = 90$$ cm, and $$f_2 = 600$$ Hz gives

$$400 \times 90 = 600 \times L_2$$

From this,

$$L_2 = \frac{400 \times 90}{600} = \frac{36000}{600} = 60 \text{ cm}$$

The answer is 60 cm.