The elongation of a wire on the surface of the earth is $$10^{-4}$$ m. The same wire of same dimensions is elongated by $$6 \times 10^{-5}$$ m on another planet. The acceleration due to gravity on the planet will be ______ m s$$^{-2}$$. (Take acceleration due to gravity on the surface of earth $$= 10$$ m s$$^{-2}$$)

The elongation of a wire on Earth is $$10^{-4}$$ m, and the same wire on another planet is elongated by $$6 \times 10^{-5}$$ m. We need to find the acceleration due to gravity on the planet.

The elongation of a wire under its own weight (or a suspended load) is given by:

$$\Delta L = \frac{F \cdot L}{A \cdot Y}$$

where $$F$$ is the applied force (weight = $$mg$$), $$L$$ is the original length, $$A$$ is the cross-sectional area, and $$Y$$ is the Young’s modulus.

Since the same wire (same $$m$$, $$L$$, $$A$$, and $$Y$$) is used on both the Earth and the planet, the only difference is the value of $$g$$. Taking the ratio of the elongations gives:

$$\frac{\Delta L_{planet}}{\Delta L_{earth}} = \frac{g_{planet}}{g_{earth}}$$

Substituting into this ratio yields:

$$g_{planet} = g_{earth} \times \frac{\Delta L_{planet}}{\Delta L_{earth}}$$

$$g_{planet} = 10 \times \frac{6 \times 10^{-5}}{10^{-4}}$$

$$g_{planet} = 10 \times \frac{6 \times 10^{-5}}{1 \times 10^{-4}}$$

$$g_{planet} = 10 \times 0.6 = 6$$ m/s$$^2$$

Therefore, the acceleration due to gravity on the planet is 6 m/s$$^2$$.