A) \[\frac{7}{3}m\]
B) \[\frac{12}{7}m\]
C) \[\frac{10}{7}m\]
D) \[\frac{9}{7}m\]
Correct Answer: B
Solution :
Let rod is placed along\[x-\]axis. Mass of element PQ of length dx situated at\[x=x\]is The CM of the element has coordinate\[(x,0,0)\] Therefore,\[x-\]coordinate of CM of the rod will be \[{{x}_{CM}}=\frac{\int_{0}^{3}{x\,dm}}{\int_{0}^{3}{dm}}\] \[=\frac{\int_{0}^{3}{x(2+x)dx}}{\int_{0}^{3}{(2+x)dx}}\] \[=\frac{\int_{0}^{3}{(2x+{{x}^{2}})dx}}{\int_{0}^{3}{(2+x)dx}}\] \[=\frac{\left[ \frac{2{{x}^{2}}}{2}+\frac{{{x}^{3}}}{3} \right]_{0}^{3}}{\left[ 2x+\frac{{{x}^{2}}}{2} \right]_{0}^{3}}\] \[=\frac{\left[ {{(3)}^{2}}+\frac{{{(3)}^{3}}}{3} \right]}{\left[ 2\times 3+\frac{{{(3)}^{3}}}{2} \right]}=\frac{12}{7}m\]You need to login to perform this action.
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