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(D) The radius of the cylindrical jar $R = 24/2 = 12\, cm$.The rise in the water level $h = 67.5\, mm = 6.75\, cm$.The volume of the spherical ball is

Vedclass · Accepted Answer

$18$

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(D) The radius of the cylindrical jar $R = 24/2 = 12\, cm$.The rise in the water level $h = 67.5\, mm = 6.75\, cm$.The volume of the spherical ball is equal to the volume of the water displaced by it.Volume of displaced water $= \pi R^2 h = \pi 	imes (12)^2 	imes 6.75\, cm^3$.Let $r$ be the radius of the spherical ball. The volume of the ball is given by $\frac{4}{3} \pi r^3$.Equating the volumes: $\frac{4}{3} \pi r^3 = \pi 	imes 144 	imes 6.75$.$r^3 = \frac{144 	imes 6.75 	imes 3}{4} = 36 	imes 6.75 	imes 3 = 108 	imes 6.75 = 729$.$r^3 = 9^3$,which implies $r = 9\, cm$.The diameter of the ball $= 2r = 2 	imes 9 = 18\, cm$.

$A$ cylindrical jar of diameter $24\, cm$ contains water to a height of $30\, cm$. $A$ spherical steel ball is dropped into the jar and the level of the water rises by $67.5\, mm$. The diameter of the ball is......$cm$.

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