A small object of uniform density rolls up a curved surface with an initial velocity $$v'.$$  It reaches upto a maximum height of $$\frac{{3{v^2}}}{{4g}}$$  with respect to the initial position. The object is

Solution :
$$\eqalign{ & {\text{As,}}\,\,v = \sqrt {\frac{{2gh}}{{1 + \frac{{{k^2}}}{{{r^2}}}}}} \cr & {\text{Given,}}\,\,h = \frac{{3{v^2}}}{{4g}} \cr & {\text{So,}}\,\,{v^2} = \frac{{2gh}}{{1 + \frac{{{k^2}}}{{{r^2}}}}} \cr & = \frac{{2g3{v^2}}}{{4g\left( {1 + \frac{{{k^2}}}{{{r^2}}}} \right)}} = \frac{{6g{v^2}}}{{4g\left( {1 + \frac{{{k^2}}}{{{r^2}}}} \right)}} \cr & 1 = \frac{3}{{2\left( {1 + \frac{{{k^2}}}{{{I^2}}}} \right)}} \cr & {\text{or}}\,\,1 + \frac{{{k^2}}}{{{r^2}}} = \frac{3}{2}\,\,{\text{or}}\,\,\frac{{{k^2}}}{{{r^2}}} = \frac{3}{2} - 1 = \frac{1}{2} \cr & {k^2} = \frac{1}{2}{r^2}\,\,\left( {{\text{Equation of disc}}} \right) \cr} $$
Hence, the object is disc.