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A body of mass ‘m’ is taken from the earth’s surface to the height equal to twice the radius (R) of the earth. The change in potential energy of body will be
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2 mgR 3 - 3 mgR
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1 mgR 3 - mg2R
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Correct Option: A
| Initial P. E., Ui = | ||
| R |
| Final P.E., Uf = | [∵ R' = R + 2R = 3R] | |
| 3R |
| ∴ Change in potential energy, ∆U = | + | |||
| 3R | R |
| = |  | 1 - |  | = | = | mgR | ||||||
| R | 3 | 3 | R | 3 |
| ∵ | = mgR | | |
| R |
| ALTERNATE : ∆U = | |||||
| 1 + | |||||
| R | |||||
| By placing the value of h = 2R we get ∆U = | mgR | |
| 3 |
