A) \[40\]
B) \[80\]
C) \[50\]
D) \[0.01\]
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 shirts to\[{{O}_{2}}\] at a distance \[{{x}_{2}}\] from original centre. The position of new centre of mass is given by \[{{X}_{CM}}=\frac{-\sigma \cdot \pi {{b}^{2}}\cdot {{x}_{1}}}{\sigma \cdot \pi {{a}^{2}}-\sigma \cdot \pi {{b}^{2}}}\] 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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