Question

A hemisphere and a solid cone have a common base. The centre of mass of the common structure coincides with the centre of the common base. If $R$ is the radius of hemisphere and $h$ is height of the cone, then:
(A) $\dfrac{h}{R} = \sqrt 3$
(B) $\dfrac{h}{R} = \dfrac{1}{{\sqrt 3 }}$
(C) $\dfrac{h}{R} = 3$
(D) $\dfrac{h}{R} = \dfrac{2}{{\sqrt 3 }}$

Answer 1

Since they have a common base, then the base area (thus the radius) of the two are the same. The density must be equal for the individual centre of mass to coincide with the centre of mass of the common structure

Formula used: In this solution we will be using the following formula;
${V_C} = \dfrac{1}{3}\pi {R^2}h$ where ${V_C}$ is the volume of a cone, $R$ is the base radius and $h$ is the height of the cone.
${V_H} = \dfrac{4}{6}\pi {R^3}$ where ${V_H}$ is the volume of a hemisphere. $R$ is the base radius of the hemisphere.
$m = \rho V$ where $m$ is the mass of a body, $\rho$ is the density, and, $V$ is the volume of the body.

Complete step by step solution:
To solve this, the mass of the individual mass must be calculated.
For the cone ${m_C} = \rho {V_C} = \rho \dfrac{1}{3}\pi {R^2}h$ where $\rho$ is the density and $\dfrac{1}{3}\pi {R^2}h$ is the volume of the cone
For the hemisphere ${m_H} = \rho {V_H} = \rho \dfrac{4}{6}\pi {R^3}$ where $\dfrac{4}{6}\pi {R^3}$ is the volume of the hemisphere.
Now the location of the centre of mass for the common structure can be given as
$y = \dfrac{{{m_C}{y_C} + {m_H}{y_H}}}{{{m_C} + {m_H}}}$ where ${y_C}$ is the centre of mass of the cone alone, and ${y_H}$ is the centre of mass of the hemisphere alone.
But by knowledge, ${y_C} = \dfrac{h}{4}$ and ${y_H} = \dfrac{{3R}}{8}$
Now, if we assume the common base is an origin and the hemisphere is at the negative $y$ – axis, then
${y_H} = - \dfrac{{3R}}{8}$ and $y = 0$
Then, inserting all known values into $y = \dfrac{{{m_C}{y_C} + {m_H}{y_H}}}{{{m_C} + {m_H}}}$ we have
$0 = \dfrac{{\rho \dfrac{1}{3}\pi {R^2}h\left( {\dfrac{h}{4}} \right) + \rho \dfrac{4}{6}\pi {R^3}\left( { - \dfrac{{3R}}{8}} \right)}}{{\rho \dfrac{1}{3}\pi {R^2}h + \rho \dfrac{4}{6}\pi {R^3}}}$
Cross multiplying we have
$\rho \dfrac{1}{3}\pi {R^2}h\left( {\dfrac{h}{4}} \right) - \rho \dfrac{4}{6}\pi {R^3}\left( {\dfrac{{3R}}{8}} \right) = 0$
$\Rightarrow \dfrac{1}{3}\pi {R^2}h\left( {\dfrac{h}{4}} \right) = \dfrac{4}{6}\pi {R^3}\left( {\dfrac{{3R}}{8}} \right)$
Cancelling common terms, and simplifying, we have
$h\left( {\dfrac{h}{4}} \right) = R\left( {\dfrac{{3R}}{4}} \right)$
Which can be simplified further to,
${h^2} = 3{R^2}$
$\Rightarrow h = R\sqrt 3$
Dividing by $R$ we get,
$\dfrac{h}{R} = \sqrt 3$
Hence, the correct option is A.

Note:
Note that alternatively, we could decide to make the cone lie on the negative axis and hemisphere on the positive axis. Hence
${y_C} = - \dfrac{h}{4}$ and ${y_H} = \dfrac{{3R}}{8}$ , and the equation becomes
$\rho \dfrac{1}{3}\pi {R^2}h\left( { - \dfrac{h}{4}} \right) + \rho \dfrac{4}{6}\pi {R^3}\left( {\dfrac{{3R}}{8}} \right) = 0$
$\Rightarrow \rho \dfrac{1}{3}\pi {R^2}h\left( {\dfrac{h}{4}} \right) = \rho \dfrac{4}{6}\pi {R^3}\left( {\dfrac{{3R}}{8}} \right)$
Which is identical to the above.