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**Hint:**Calculate the CM for each of the mentioned masses from the method of integrating the mass and finding the center of mass. Take an infinitesimal element correspondingly and proceed accordingly.

**Complete step by step solution:**

For CM of a uniform semicircular ring, let us assume that the ring is kept in the $y>0$ region. Let us consider $dm$ at an angle $\theta $ from the positive x-axis, which subtends an angle $d\theta $ at the origin. The length of this $dm$ is $dl=Rd\theta$.

Length of a semicircular ring, $l=\pi R$

If mass for length $l=M$, then mass for length $dl=\dfrac{M}{l}dl$.

$\Rightarrow dm=\dfrac{M}{l}Rd\theta $

Therefore, $C{{M}_{x}}=\dfrac{1}{M}\int\limits_{0}^{\pi }{R\cos \theta \dfrac{M}{\pi R}}Rd\theta $

$\Rightarrow C{{M}_{x}}=0$

And, $C{{M}_{y}}=\dfrac{1}{M}\int\limits_{0}^{\pi }{R\sin \theta \dfrac{M}{\pi R}}Rd\theta $

$\Rightarrow C{{M}_{y}}=\dfrac{2R}{\pi }$

For CM of a semicircular disc, let us assume that the disc is kept in the $y>0$ region. Let us consider a ring on the disc of radius $r$, of thickness $dr$ and mass $dm$.

Mass per unit area of the disc:

$\dfrac{M}{A}=\dfrac{M}{\left( \dfrac{\pi {{R}^{2}}}{2} \right)}=\dfrac{2M}{\pi {{R}^{2}}}$

Area of the considered ring:

${{A}_{r}}=\pi rdr$

Therefore, $dm=\dfrac{M}{A}{{A}_{r}}=\dfrac{2M}{\pi {{R}^{2}}}\pi rdr=\dfrac{2M}{{{R}^{2}}}rdr$

Hence, $C{{M}_{y}}=\dfrac{1}{M}\int\limits_{0}^{M}{ydm=\dfrac{1}{M}\int\limits_{0}^{R}{\dfrac{2r}{\pi }}\dfrac{2M}{{{R}^{2}}}}rdr$

$C{{M}_{y}}=\dfrac{4}{\pi {{R}^{2}}}\int\limits_{0}^{R}{{{r}^{2}}}dr=\dfrac{4R}{3\pi }$

For CM of solid Hemisphere, let us assume that the hemisphere is kept in the $y>0$ region. Let us assume a disc contained inside the hemisphere at a height $y$, of thickness $dy$ and mass $dm$, with radius $r$.

Mass per unit volume of the hemisphere:

$\dfrac{M}{V}=\dfrac{M}{\left( \dfrac{2}{3}\pi {{R}^{3}} \right)}=\dfrac{3M}{2\pi {{R}^{3}}}$

Let the volume of the disc considered be $dV$.

$dV=\pi {{r}^{2}}dy$

By applying Pythagoras theorem:

${{y}^{2}}+{{r}^{2}}={{R}^{2}}$ $\Rightarrow {{r}^{2}}={{R}^{2}}-{{y}^{2}}$

Therefore, $dV=\pi \left( {{R}^{2}}-{{y}^{2}} \right)dy$

$dm=\dfrac{M}{V}dV=\dfrac{3M}{2\pi {{R}^{2}}}\pi ({{R}^{2}}-{{y}^{2}})dy$

$C{{M}_{y}}=\dfrac{1}{M}\int\limits_{0}^{M}{ydm}=\dfrac{1}{M}\int\limits_{0}^{R}{y}\dfrac{3M}{2{{R}^{3}}}\left( {{R}^{2}}-{{y}^{2}} \right)dy$

$C{{M}_{y}}=\dfrac{3}{2{{R}^{3}}}\int\limits_{0}^{R}{(y{{R}^{2}}-{{y}^{3}})}dy$

$C{{M}_{y}}=\dfrac{3R}{8}$

For the CM of a hemispherical shell, let us assume that the hemispherical shell is kept in the region $y>0$. Let us consider a ring contained at the periphery of the shell at height $y$, radius $r$ and subtends and $d\theta$ at the origin, $\theta$ is the angle at which the ring is considered from the x-axis. Therefore, the thickness of the ring is $Rd\theta$. Let the mass of this ring be $dm$ .

Let $dA$ be the surface area of the ring.

$dA=2\pi rRd\theta $

From trigonometry, it can be easily proved that $r=R\cos \theta $ and $y=R\sin \theta $ .

$\Rightarrow dA=2\pi R\cos \theta Rd\theta =2\pi {{R}^{2}}\cos \theta d\theta $

Mass per unit area of the hemispherical shell:

$\dfrac{M}{A}=\dfrac{M}{2\pi {{R}^{2}}}$

$dm=\dfrac{M}{2\pi {{R}^{2}}}2\pi {{R}^{2}}\cos \theta d\theta =M\cos \theta d\theta $

$C{{M}_{y}}=\dfrac{1}{M}\int\limits_{0}^{M}{ydm}=\dfrac{1}{M}\int\limits_{0}^{\dfrac{\pi }{2}}{R\sin \theta M\cos \theta d\theta }$

$C{{M}_{y}}=\dfrac{R}{2}\int\limits_{0}^{\dfrac{\pi }{2}}{\sin (2\theta )d\theta }$

$C{{M}_{y}}=\dfrac{R}{2}$

Therefore, only statement 4 is correct.

**Hence, option (C) is correct.**

**Note:**Do the integration very carefully as many errors can be done while doing integration. Don’t forget to write constants while doing integration. Take proper care of the limits of the integration. Do the manipulation(s) carefully, if required.

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