Directions : In the following question more than one of the answers given may be correct. Select the correct answer and mark it according to the code:
A particle of mass m is moved from the surface of the earth to a height h. The work done by an external agency to do this is (1) \[mgh\text{ }for\text{ }h<<R\] (2) \[mgh\] for all h (3) \[\frac{1}{2}\text{mgh}\,for\text{ }h=R\] (4)\[-\frac{1}{2}\text{mgh}\,for\text{ }h=R\]A) 1, 2 and 3 are correct
B) 1 and 2 are correct
C) 2 and 4 are correct
D) 1 and 3 are correct
Correct Answer: D
Solution :
\[{{U}_{i}}=-\frac{GMm}{R}\] \[{{U}_{f}}=-\frac{GMm}{(R+h)}\] Work done = change in gravitational potential energy \[E={{U}_{f}}-{{U}_{i}}\] \[=-GMm\left[ \frac{1}{R+h}-\frac{1}{R} \right]\] \[W=-\frac{GMm}{R}\left[ {{\left( 1+\frac{h}{R} \right)}^{-1}}-1 \right]\] \[W=-\frac{GMm}{R}\left[ 1-\frac{h}{R}-1 \right]\] \[W=\frac{GMmh}{{{R}^{2}}}\] \[W=mgh\]for \[h\le R\] Also\[{{U}_{i}}=-\frac{GMm}{R}\]and on the surface \[h=R\] \[{{U}_{f}}=-\frac{GMm}{R+R}\] \[=-\frac{GMm}{2R}\] Work done\[(W)={{U}_{f}}-{{U}_{i}}\] \[=\frac{GMm}{2R}\] \[W=\frac{1}{2}mgR\] \[\left[ \because g=\frac{GM}{{{R}^{2}}} \right]\]You need to login to perform this action.
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