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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Gravitation
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A rocket is launched with velocity 10 km/s if radius of earth is R, then maximum height attained by it will be

\[(a)\;2\; R \quad (b)\;4\;R \quad (c)\; 3\;R \quad (d)\;5\;R\]
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$KE$ given =$P.E$ of the body
$\large\frac{1}{2} $$mv^2=mg (R+h)$
Where $'h'$ maximum height reached by the satellite
$\bigg( 10 \times \large\frac{5}{8}\bigg)^2 $$\times 10^6= 2\times 10(R+h)$
Solving we get
Where $R= 6.4 \times 10^6 \;m$
$\quad h=4R$
Hence b is the correct answer. 


answered Aug 10, 2013 by meena.p
edited Feb 17, 2014 by meena.p

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