Thank you for reporting, we will resolve it shortly
Q.
The percentage decrease in the weight of a rocket, when taken to a height of $32 \,km$ above the surface of earth will, be :
(Radius of earth $=6400 \,km$ )
Acceleration due to gravity at a height $h << R$ is
$ g ^{\prime}= g \left(1-\frac{2 h }{ R }\right) $
$ \therefore \frac{\Delta g }{ g }=\frac{2 h }{ R } $
$ \Rightarrow \frac{\Delta g }{ g } \times 100=\frac{2 h }{ R } \times 100 $
$=2 \times \frac{32}{6400} \times 100=1 \%$