SSC JE Mechanical Engineering 2013 Paper

Centre of pressure for inclined plane submerged in liquid is given by
$\mathit{h}=\frac{{\mathit{I}}_{\mathrm{cg}}.{\sin }^2\mathit{\theta }}{\mathit{A}\stackrel{-}{\mathit{x}}}+\stackrel{-}{\mathit{x}}$${\mathit{I}}_{\mathit{G}}=\frac{\mathit{b}{\mathit{d}}^3}{12}$
h = centre of pressure
${\mathit{I}}_{\mathit{G}}=$ moment of inertia of plane area
$\stackrel{-}{\mathit{x}}=$ centre of gravity.
Q = angle of inclination,
A = area of plane.
∴ Vertical distance of C.P. below C.G.
$\mathit{h}-\stackrel{-}{\mathit{x}}=\frac{{\mathit{I}}_{\mathrm{cg}}.{\sin }^2\mathit{\theta }}{\mathit{A}\stackrel{-}{\mathit{x}}}$