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We can draw the diagram

Given that, Volume of cylinder $= 9\pi$

Let the radius of the hollow cylinder be $r$ cm.

$\Rightarrow \pi r^{2}h=9 \pi$

$\Rightarrow r^{2}(3)=9$

$\Rightarrow r^{2}=3$

$\Rightarrow \boxed{r=\sqrt{3}\;\text{cm}}$

The diameter of ball $= 4\;\text{cm}$

Let the radius of the ball be $R$ cm.

The radius of the ball $R= 2\;\text{cm}$

$\triangle \text{OPQ}$ is a right-angle- triangle. We can apply the Pythagorean theorem.

$\text{(OQ)}^{2} = \text{(OP)}^{2} + \text{(PQ)}^{2}$

$\Rightarrow 2^{2}= \text{(OP)}^{2} + (\sqrt{3})^{2}$

$\Rightarrow \text{(OP)}^{2} = 4-3 $

$\Rightarrow \text{(OP)}^{2}=1$

$\Rightarrow \boxed{\text{OP} = 1\;\text{cm}}$

$\therefore$ The vertical distance of the topmost point of the ball from the base of the cylinder

$ \qquad = h + \text{OP} +R = 3+1+2 = 6 \;\text{cm}.$

Correct Answer $:\text{D}$