c
(c)
Points of equilibrium of the spring will be when no force acts on it.
$k x=\left(m_1+m_2\right) g$
$x=\frac{\left(m_1+m_2\right) g}{k}$
The new equilibrium position which will be the mean position of $S.H.M.$ will be simply $\frac{m_2 g}{k}$
New amplitude will be maximum displacement from $\frac{m_2 g}{k}$ which is :
$A=\frac{\left(m_1+m_2\right) g}{k}-\frac{m_2 g}{k}$
or $A=\frac{m_1 g}{k}$
or $A=\frac{1 \times 10}{12.5}$
or $A=\frac{4}{5} \,m$
$\therefore A=0.8 \,m$ or $80 \,cm$