AP EAMCET · PHYSICS · Gravitation
A body of mass \(m\) is placed on the earth's surface. It is taken from the earth's surface to a height \(h=3 R,(R\) is radius of earth \()\). The change in gravitational potential energy of the body is
- A \(\left(\frac{2}{3}\right) m g R\)
- B \(\left(\frac{3}{4}\right) m g R\)
- C \(\left(\frac{1}{2}\right) m g R\)
- D \(\left(\frac{1}{4}\right) m g R\)
Answer & Solution
Correct Answer
(B) \(\left(\frac{3}{4}\right) m g R\)
Step-by-step Solution
Detailed explanation
Gravitational potential energy on the surface of earth, \[ U_1=-\frac{G M m}{R} \] where, \(M=\) mass of earth. and \(\quad R=\) radius of earth. Gravitational potential energy at height \(h=3 R\) is given as…
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