A) 66.7 %
B) 33.3 %
C) 75 %
D) 20.3 %
Correct Answer: B
Solution :
[b] \[{{E}_{1}}=-\frac{GMm}{2r}\] \[{{E}_{2}}=-\frac{GMm}{2\times \frac{3}{2}\times r}=-\frac{GMm}{3r}\] % increase in energy \[=-\frac{\frac{GMm}{3r}+\frac{GMm}{2r}}{-\frac{GMm}{2r}}\times 100\] \[=-\frac{\frac{GMm}{6r}}{\frac{-GMm}{2r}}=\frac{100}{3}%\] = 33.33 %You need to login to perform this action.
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