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The percentage decrease in the weight of a rocket, when taken to a height of $32 \mathrm{~km}$ above the surface of earth will, be :

$\mathrm{\text { (Radius of earth }=6400 \mathrm{~km})}$
Option: 1
$1 \%$

Option: 2
$3 \%$

Option: 3
$4 \%$

Option: 4
$0.5 \%$

Answers (1)

best_answer

At,

$\mathrm{h=32\: km}$

$\mathrm{g_{n}=g(1-\frac{2h}{R})}$

$\mathrm{=g(1-\frac{64}{6400})}$

$\mathrm{=g(1-\frac{64}{6400})}$

$\mathrm{g_{n}=\frac{99}{100}g}$

Weight at height $\mathrm{h=mg_{n}}$

$\mathrm{=\frac{99}{100}mg}$

Weight on earth surface = $\mathrm{mg}$

Percentage decrease in the weight

$=\mathrm{\left ( \frac{mg-mg_{n}}{mg} \right )\times 100}$

$=1%$

Hence (1) is correct option

Posted by

seema garhwal

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