A) \[-2mg{{\operatorname{R}}_{e}}\]
B) \[2mg{{\operatorname{R}}_{e}}\]
C) \[\frac{1}{2}mg{{\operatorname{R}}_{e}}\]
D) \[-\frac{1}{2}mg{{\operatorname{R}}_{e}}\]
Correct Answer: D
Solution :
\[\Delta U={{U}_{2}}-{{U}_{1}}\] \[=\frac{mgh}{1+\frac{h}{{{\operatorname{R}}_{e}}}}=\frac{mg{{\operatorname{R}}_{e}}}{1+\frac{{{\operatorname{R}}_{e}}}{{{\operatorname{R}}_{e}}}}=\frac{mg{{\operatorname{R}}_{e}}}{2}\] \[\therefore \]\[{{U}_{2}}=\Delta U+{{U}_{1}}\] \[{{U}_{2}}=\frac{mg{{\operatorname{R}}_{e}}}{2}-mg{{\operatorname{R}}_{e}}\] \[\Rightarrow \] \[{{U}_{2}}=-\frac{1}{2}mg{{\operatorname{R}}_{e}}\]You need to login to perform this action.
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