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As shown in fig. when a spherical cavity (centred at O) of radius 1 is cut out of a uniform sphere of radius R (centred at C), the centre of mass of the remaining (shaded) part of the sphere is at G, i.e on the surface of the cavity. R can be determined by the equation :
Option: 1 \left ( R^{2} + R + 1 \right ) \left ( 2 - R \right ) = 1
Option: 2 \left ( R^{2} + R - 1 \right ) \left ( 2 - R \right ) = 1
Option: 3 \left ( R^{2} - R - 1 \right ) \left ( 2 - R \right ) = 1
Option: 4 \left ( R^{2} - R + 1 \right ) \left ( 2 - R \right ) = 1
 

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best_answer

 

 

 

Taking centre at O for the cavity and the Center of the solid sphere at C.
Let origin be at G
Mass of the sphere is taken to be \mathrm{m}_{\mathrm{s}}, which can be given as:

\mathrm{m}_{\mathrm{s}}=\left(\frac{4}{3} \pi \mathrm{R}^{3} \cdot \rho\right)\mathrm{m}_{\mathrm{s}}=\left(\frac{4}{3} \pi \mathrm{R}^{3} \rho\right)

Mass of cavity is taken to be \mathrm{m}_{\mathrm{c}}, which can be given as:

\mathrm{m}_{\mathrm{c}}=-\left(\frac{4}{3} \pi \times 1^{2} . \rho\right)

where \rho is the density of the material.

Center of mass of sphere:


\begin{aligned} &\mathrm{x}_{8}=\mathrm{GC} \quad\left[\mathrm{x}_{\mathrm{s}} \text { can be given as } \mathrm{x}_{\mathrm{c}}-\mathrm{x}_{\mathrm{G}}=\mathrm{GC}\right] \\ &\mathrm{x}_{\mathrm{s}}=2-\mathrm{R} \end{aligned}

Center of mass of cavity:

\begin{aligned} &x_{c}=2 \times 1-1 \\ &x_{c}=1 \end{aligned}


\mathrm{x}_{\mathrm{com}}=\frac{\mathrm{m}_{\mathrm{s}} \mathrm{x}_{\mathrm{s}}+\mathrm{m}_{\mathrm{c}} \mathrm{x}_{\mathrm{c}}}{\mathrm{m}_{\mathrm{x}}+\mathrm{m}_{\mathrm{c}}}=\frac{\left(\frac{4}{3} \pi \mathrm{R}^{3} \rho\right) \times(2-\mathrm{R})+\left(-\frac{4}{3} \pi \rho\right) \times 1}{\frac{4 \pi}{3} \pi \mathrm{R}^{3} \rho+\left(-\frac{4}{3} \pi \rho\right)}

Therefore the centre of mass is at G, \mathrm{x}_{ \mathrm{com}}=0

\begin{aligned} &\Rightarrow 0=\left(\frac{4}{3} \pi R^{3} \rho\right)(2-R)-\frac{4}{3} \pi \rho \\ &0=R^{3}(2-R)-1 \Rightarrow 2 R^{3}-R^{4}-1=0 \\ &R^{3}+R^{3}-R^{4}-1=0 \\ &\Rightarrow R^{3}-R^{3}(R-1)=1 \\ &\Rightarrow(R-1)\left(R^{2}+1+R\right)-R^{3}(R-1)=0 \\ &\Rightarrow(R-1)\left(R^{2}+1+R-R^{3}\right)=0 \\ &\Rightarrow R^{2}+1+R-R^{3}=0 \end{aligned}

\begin{aligned} &\Rightarrow \mathrm{R}^{2}+\mathrm{R}-\mathrm{R}^{3}+2=1 \\ &\Rightarrow 2 \mathrm{R}^{2}-\mathrm{R}^{2}+2 \mathrm{R}-\mathrm{R}-\mathrm{R}^{3}+2=1 \\ &\Rightarrow 2 \mathrm{R}^{2}-\mathrm{R}^{3}-\mathrm{R}^{2}+2 \mathrm{R}-\mathrm{R}+2=1 \\ &\Rightarrow \mathrm{R}^{2}(2-\mathrm{R})+\mathrm{R}(2-\mathrm{R})+1(2-\mathrm{R})=1 \\ &\Rightarrow(2-\mathrm{R})\left(\mathrm{R}^{2}+\mathrm{R}+1\right)=1 \end{aligned}

Posted by

vishal kumar

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