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The depth of centre of pressure (h) for a vertically immersed surface from the liquid surface is given by (where $${I_{\text{G}}}$$ = Moment of inertia of the immersed surface about horizontal axis through its centre of gravity, A = Area of immersed surface and x = Depth of centre of gravity of the immersed surface from the liquid surface)

The depth of centre of pressure (h) for a vertically immersed surface from the liquid surface is given by (where $${I_{\text{G}}}$$ = Moment of inertia of the immersed surface about horizontal axis through its centre of gravity, A = Area of immersed surface and x = Depth of centre of gravity of the immersed surface from the liquid surface) Correct Answer $$\frac{{{I_{\text{G}}}}}{{{\text{A}}\overline {\text{x}} }} + \overline {\text{x}} $$

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The depth of centre of pressure (h) for a vertically immersed surface from the liquid surface is given by (where IG = Moment of inertia of the immersed surface about horizontal axis through its centre of gravity, A = Area of immersed surface, and x = Depth of centre of gravity of the immersed surface from the liquid surface)
According to parallel axis theorem, the moment of inertia of a section about an axis parallel to the axis through center of gravity (i.e. $$I$$P) is given by (where, A = Area of the section, $$I$$G = Moment of inertia of the section about an axis passing through its C.G. and h = Distance between C.G. and the parallel axis.)
According to parallel axis theorem, the moment of inertia of a section about an axis parallel to the axis through center of gravity (i.e. $${I_{\text{P}}}$$) is given by (where, A = Area of the section, $${I_{\text{G}}}$$ = Moment of inertia of the section about an axis passing through its C.G. and h = Distance between C.G. and the parallel axis.)
The location of centre of pressure, which defines the point of application of the total pressure force on the surface, can be calculated by applying the principle of moments according to which "sum of the moment of the resultant force about an axis is equal to the sum of the components about the same axis". The centre of pressure of a rectangular surface (of width 'w') immersed vertically in a static mass of fluid is at a depth of(where, y = depth of the liquid)
The location of centre of pressure, which defines the point of application of the total pressure force on the surface, can be calculated by applying the principle of moments according to which "sum of the moment of the resultant force about an axis is equal to the sum of the components about the same axis". The centre of pressure of a rectangular surface (of width 'w') immersed vertically in a static mass of fluid is at a depth of (where, y = depth of the liquid)
The moment of inertia of a uniform sphere of radius, r about an axis passing through its centre is given by $$\frac{2}{5}\left( {\frac{{4\pi }}{3}{r^5}\rho } \right).$$    A rigid sphere of uniform mass density $$\rho $$ and radius R has two smaller spheres of radii $$\frac{R}{2}$$ hollowed out of it as shown in the figure given below.
Classical Mechanics mcq question image
The moment of inertia of the resulting body about Y-axis is
The total pressure on a horizontally immersed surface is (where w = Specific weight of the liquid, A = Area of the immersed surface, and x = Depth of the centre of gravity of the immersed surface from the liquid surface)
The total pressure on a horizontally immersed surface is (where w = Specific weight of the liquid, A = Area of the immersed surface and x = Depth of the centre of gravity of the immersed surface from the liquid surface)
A rectangle is having base b and height h. The ratio of the moment of inertia when axis is passing through its base to moment of inertia of when axis passing through center of gravity and parallel to base is ___________
In order to have a connecting rod equally strong in buckling about X-axis and Y-axis, the moment of inertia about X-axis should be __________ the moment of inertia about Y-axis.