A) \[a=2b\]
B) \[2a=b\]
C) \[a=b\]
D) \[3a=2b\]
Correct Answer: B
Solution :
Centre of mass of the rod is given by: \[{{x}_{cm}}=\frac{\int\limits_{0}^{L}{(ax+\frac{b{{x}^{2}}}{L})dx}}{\int\limits_{0}^{L}{(a+\frac{bx}{L})dx}}\] \[=\frac{\frac{a{{L}^{2}}}{2}+\frac{b{{L}^{2}}}{3}}{aL+\frac{bL}{2}}=\frac{L\left( \frac{a}{2}+\frac{b}{3} \right)}{a+\frac{b}{2}}\] Now \[\frac{7L}{12}=\frac{\frac{a}{2}+\frac{b}{3}}{a+\frac{b}{2}}\] On solving we get, \[b=2a\]You need to login to perform this action.
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