A) 0.4 cm
B) 2.4 cm
C) 1.8 cm
D) 1.2 cm
Correct Answer: A
Solution :
The situation can be shown as: Let radius of complete disc is a and that of small disc is b. Also let centre of mass now shifts to \[{{\text{O}}_{\text{2}}}\] at a distance \[{{x}_{2}}\]from original centre. The position of new centre of mass is given by \[{{X}_{CM}}=\frac{-\sigma .\pi {{b}^{2}}.{{x}_{1}}}{\sigma .\pi {{a}^{2}}-\sigma .\pi {{b}^{2}}}\] Here, \[a=6\,cm,\,b=2\,cm,{{x}_{1}}=3.2\,cm\] Hence,\[{{X}_{CM}}=\frac{-\sigma \times \pi {{(2)}^{2}}\times 3.2}{\sigma \times \pi \times {{(6)}^{2}}-\sigma \times \pi \times {{(2)}^{2}}}\] \[=\frac{12.8\,\pi }{32\,\pi }\] \[=-0.4\,cm\]You need to login to perform this action.
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