A hollow tilted cylindrical vessel of negligi | vedclass

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(C) The cylinder will topple when the centre of mass of the filled water column lies outside the right edge of the base. Since the water forms a cylin

Vedclass · Accepted Answer

$h=a 	an 	heta$

Vedclass · Answer

(C) The cylinder will topple when the centre of mass of the filled water column lies outside the right edge of the base. Since the water forms a cylindrical shape of height $h$ (measured vertically),the centre of mass of this water column lies at its geometric centre.Let the base of the cylinder be on the horizontal plane. The right edge of the base is at point $A$. The centre of mass of the water column is at a horizontal distance of $a/2$ from the left edge,or more simply,the vertical line passing through the centre of mass must pass through the base for stability.From the geometry of the tilted cylinder,the horizontal distance from the right edge of the base to the vertical line passing through the centre of the water column is $a/2$. The vertical height of the centre of mass of the water column is $h/2$.For the cylinder to be on the verge of toppling,the vertical line passing through the centre of mass must pass through the right edge of the base. Considering the triangle formed by the tilt angle $	heta$,the horizontal distance from the centre of the base to the right edge is $a/2$. The vertical height of the centre of mass is $h/2$. From the geometry,$	an 	heta = \frac{	ext{vertical height}}{	ext{horizontal distance}} = \frac{h/2}{a/2} = \frac{h}{a}$.Therefore,$h = a 	an 	heta$.

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