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Q.

A uniform solid right circular cone of base radius r is joined to a uniform solid hemisphere of radius r and of the same density, so as to have a common face. The centre of mass of the composite solid lies on the common face. The height of the cone is

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a

4r

b

3r

c

2r

d

$r \sqrt{6}$

answer is C.

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Detailed Solution

Mass of hemisphere $= \frac{2}{3} {πr}^{3} ρ = m_{1}$
Mass of cone $= \frac{1}{3} π^{2} rhρ = m_{2}$
Mass of moments about CM is zero so
$\begin{array}{l} m_{1} (\frac{r}{2}) = m_{2} (\frac{h}{4}) \\ \Rightarrow \frac{2}{3} {πr}^{3} ρ (\frac{r}{2}) = \frac{{πr}^{2} h}{3} ρ (\frac{h}{4}) \\ \Rightarrow h = 2 r \end{array}$

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