A) \[\left( {{R}^{2}}\text{+}R+1 \right)\left( 2R \right)=1\]
B) \[\left( {{R}^{2}}R+1 \right)\left( 2-R \right)=1\]
C) \[\left( {{R}^{2}}R1 \right)\left( 2R \right)\,\,\text{=}\,1\]
D) \[\left( {{R}^{2}}+R1 \right)\left( 2R \right)=1\]
Correct Answer: A
Solution :
\[{{M}_{0}}=\frac{4}{3}\pi {{R}^{3}}\rho \] \[{{M}_{cavity}}=\frac{4}{3}\pi {{\left( 1 \right)}^{3}}\rho \] \[{{M}_{\left( \operatorname{Re}maining \right)}}=\frac{4}{3}\pi {{R}^{3}}\rho -\frac{4}{3}\pi {{\left( 1 \right)}^{3}}\rho \] \[\therefore \left( \frac{4}{3}\pi {{R}^{3}}\rho -\frac{4}{3}\pi {{\left( 1 \right)}^{3}}\rho \right)\times 2+\frac{4}{3}\pi {{\left( 1 \right)}^{3}}\rho =\frac{4}{3}\pi {{R}^{3}}\rho \cdot R\] \[\Rightarrow \,\,\,\,\,{{R}^{4}}-2{{R}^{3}}+1=0\] \[\because \,\,\,\,\,\,\,R\ne 1\] \[\therefore \,\,\,\,\,\,\,{{R}^{3}}-{{R}^{2}}-R-1=0\] \[\Rightarrow \,\,\left( {{R}^{2}}+R+1 \right)\left( 2-R \right)=1\]You need to login to perform this action.
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