A) 0.4cm
B) 2.4cm
C) 1.8cm
D) 1.2cm
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 \[{{O}_{2}}\] 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, \[{{X}_{CM}}=\frac{-\sigma \times \pi {{(2)}^{2}}\times 3.2}{\sigma \times \pi \times {{(6)}^{2}}-\sigma \times \pi \times {{(2)}^{2}}}\] \[=-\frac{12.8}{32\,\pi }\] = - 0.4 cmYou need to login to perform this action.
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