### Neet Solved Paper 2013 Question 10

##### Question: A body of mass m taken from the earth’s surface to the height equal to twice the radius (R) of the earth. The change is potential energy of body will be

#### Options:

A) $ mg2R $

B) $ \frac{2}{3}mgR $

C) $ 3mgR $

D) $ \frac{1}{3}mgR $

## Show Answer

#### Answer:

Correct Answer: B

#### Solution:

Change in potential energy $ \Delta U=-\frac{GMm}{R+2R}-( -\frac{GMm}{R} ) $

$ =-\frac{GMm}{3R}+\frac{GMm}{R} $ $ =\frac{2GMm}{3R}=\frac{2}{3}mgR $ $ [ \because g=\frac{GM}{R^{2}} ] $