If the gravitational field in the space is given as $$-\frac{K}{r^2}$$. Taking the reference point to be at $$r = 2$$ cm with gravitational potential $$V = 10$$ J kg$$^{-1}$$. Find the gravitational potentials at $$r = 3$$ cm in SI unit (Given, that $$K = 6$$ J cm kg$$^{-1}$$)

We need to find the gravitational potential at $$r = 3$$ cm, given the gravitational field $$g = -\frac{K}{r^2}$$ and $$V(r=2) = 10$$ J/kg.

Relation between field and potential:

$$V(r) - V(r_0) = -\int_{r_0}^{r} g \, dr$$

Calculating the potential difference:

$$V(3) - V(2) = -\int_{2}^{3} \left(-\frac{K}{r^2}\right) dr = K\int_{2}^{3} \frac{dr}{r^2}$$

$$= K\left[-\frac{1}{r}\right]_{2}^{3} = K\left(-\frac{1}{3} + \frac{1}{2}\right) = K \times \frac{1}{6}$$

Substituting values:

$$K = 6$$ J cm/kg (note: distances are in cm)

$$V(3) = V(2) + \frac{K}{6} = 10 + \frac{6}{6} = 10 + 1 = 11$$ J/kg

The gravitational potential at $$r = 3$$ cm is $$11$$ J/kg.

The correct answer is Option 2: $$11$$.