The percentage decrease in the weight of a rocket, when taken to a height of $$32 \text{ km}$$ above the surface of earth will be (Radius of earth $$= 6400 \text{ km}$$)

We need to find the percentage decrease in weight when a rocket is taken to a height of $$h = 32 \text{ km}$$ above the Earth's surface. The radius of Earth is $$R = 6400 \text{ km}$$.

We start by expressing the acceleration due to gravity at height $$h$$ above the surface as:

$$g' = g\left(\frac{R}{R + h}\right)^2$$

Since $$h \ll R$$ ($$32 \text{ km} \ll 6400 \text{ km}$$), we can use the binomial approximation:

$$g' \approx g\left(1 - \frac{2h}{R}\right)$$

Next, the fractional decrease in gravitational acceleration is given by:

$$\frac{\Delta g}{g} = \frac{g - g'}{g} = \frac{2h}{R}$$

Substituting the values, we get $$\frac{\Delta g}{g} = \frac{2 \times 32}{6400} = \frac{64}{6400} = \frac{1}{100} = 0.01$$

Finally, converting this to a percentage gives:

$$\text{Percentage decrease} = \frac{\Delta g}{g} \times 100 = 0.01 \times 100 = 1\%$$

Since weight is directly proportional to $$g$$, the percentage decrease in weight is also $$1\%$$.

The correct answer is Option A: $$1\%$$.