The density and breaking stress of a wire are $$6 \times 10^4 \text{ kg/m}^3$$ and $$1.2 \times 10^8 \text{ N/m}^2$$ respectively. The wire is suspended from a rigid support on a planet where acceleration due to gravity is $$\frac{1}{3}^{rd}$$ of the value on the surface of earth. The maximum length of the wire with breaking is ______ m (take, $$g = 10 \text{ m/s}^2$$).

The density is $$\rho = 6 \times 10^4$$ kg/m$$^3$$, the breaking stress is $$\sigma = 1.2 \times 10^8$$ N/m$$^2$$, and the effective acceleration is $$g' = g/3 = 10/3$$ m/s$$^2$$. The wire will break under its own weight when the stress at its top equals the breaking stress.

$$ \sigma = \frac{F}{A} = \frac{\rho A L g'}{A} = \rho L g' $$

At the breaking point, the length L is given by

$$ L = \frac{\sigma}{\rho g'} = \frac{1.2 \times 10^8}{6 \times 10^4 \times \frac{10}{3}} $$

$$ = \frac{1.2 \times 10^8}{2 \times 10^5} = 600 \text{ m} $$

Therefore, the maximum length of the wire is 600 m.