An object is taken to a height above the surface of earth at a distance $$\dfrac{5}{4}R$$ from the centre of the earth. Where radius of earth, $$R = 6400 \text{ km}$$. The percentage decrease in the weight of the object will be

An object is at a distance $$\frac{5}{4}R$$ from the centre of the Earth, i.e., at height $$h = \frac{5R}{4} - R = \frac{R}{4}$$ above the surface.

Find the acceleration due to gravity at this height.

Since the object is above the surface, at distance $$r = \frac{5R}{4}$$ from the centre:

$$g' = g\left(\frac{R}{r}\right)^2 = g\left(\frac{R}{\frac{5R}{4}}\right)^2 = g\left(\frac{4}{5}\right)^2 = \frac{16g}{25}$$

Calculate the percentage decrease in weight.

$$\text{Percentage decrease} = \frac{W - W'}{W} \times 100 = \frac{g - g'}{g} \times 100$$

$$= \left(1 - \frac{16}{25}\right) \times 100 = \frac{9}{25} \times 100 = 36\%$$

The correct answer is Option A: $$36\%$$.