A ball of mass m is dropped from a height h equal to the radius of the earth above the tunnel dug through the earth as shown in the figure. Choose the wrong options.
Answer & explanation
Correct answer: option 2
Particle performs SHM inside tunnel but not outside surface, so, if just performs periodic motion with amplitude, (R + h) :
\(\frac{1}{2} mv^2 = \frac{3GMm}{2R} - \frac{GMm}{2R}\)
\(\frac{1}{2} mv^2 = \frac{GMm}{R}\)
\(\Rightarrow v = \sqrt{\frac{2GM}{R}}\)