If one end of vertical spring is connected to the ground and other end is connected to horizontal platform at rest.

Answer: (a) 120 mg (in N/m, with \(m\) in kg and \(g\) in m/s²)

Treat the platform as light. The ball falls a height \(h\) onto the platform and then moves down a further \(h/2\) while compressing the spring, and comes momentarily to rest. Total loss of gravitational PE = PE stored in the spring:

\[mg\left(h + \frac{h}{2}\right) = \frac{1}{2}k\left(\frac{h}{2}\right)^{2}\]

\[\frac{3mgh}{2} = \frac{kh^{2}}{8}\ \Rightarrow\ k = \frac{12mg}{h}\]

With \(h = 10\ \mathrm{cm} = 0.1\ \mathrm{m}\):

\[k = \frac{12mg}{0.1} = 120\,mg\]

A common mistake is to use only \(mgh\) and forget that the ball also loses PE while compressing the spring by \(h/2\); that gives \(80\,mg\), which is not among the options.