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  • The percentage decrease in the weight of a rocket, when taken to a height of  above the surface of earth will, b

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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