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An object of mass 1 \mathrm{~kg} is taken to a height from the surface of earth which is equal to three times the radius of earth. The gain in potential energy of the object will be [If, \mathrm{g}=10 \mathrm{~ms}^{-2} and radius of earth =6400 \mathrm{~km} ]
 

Option: 1

48 \mathrm{MJ}


Option: 2

24 \mathrm{MJ}


Option: 3

36 \mathrm{MJ}


Option: 4

12 \mathrm{MJ}


Answers (1)

best_answer

\mathrm{\text { Initial } P .E . =P E_i=-\frac{G M m}{R} }

\mathrm{\text { Final PE. } =P E_f=\frac{-G M m}{(R+h)}=\frac{-G M m}{4 R} }

\mathrm{\Delta P E =PE_f-P E_i }

\mathrm{=\frac{3}{4} \frac{G M m}{R} }

\mathrm{=\frac{3 \mathrm{mgR}}{4}=\frac{3}{4} \times 1 \times 10 \times 6400\times 10^{3} }

\mathrm{\triangle P E =48 \mathrm{MJ}}

 

Posted by

avinash.dongre

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