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A small ball of mass $'m'$ is released at a height $'R'$ above the Earth surface, as shown in the figure. If the maximum depth of the ball to which it goes is $R/2$ inside the Earth through a narrow grove before coming to rest momentarily. The grove, contain an ideal spring of spring constant $K$ and natural length $R,$ the value of $K$ is ( $R$ is radius of Earth and $M$ mass of Earth)
$\frac {3GMm}{R^3}$
$\frac {6GMm}{R^3}$
$\frac {9GMm}{R^3}$
$\frac {7GMm}{R^3}$
Solution
By energy conservation
$K_{i}+U_{i}=K_{f}+U_{f}$
$0-\frac{G M m}{2 R}=\frac{1}{2} K\left(\frac{R}{2}\right)^{2} \frac{11 G M m}{8 R}$
$\frac{G M m}{R}\left[\frac{11}{8}-\frac{1}{2}\right]=\frac{1}{2} K\left(\frac{R^{2}}{4}\right)$
$\frac{7 G M m}{8 R}=\frac{K R^{2}}{8}$
$K=\frac{7 G M m}{R^{3}}$
