A) \[\frac{3}{2}\left( \frac{a+b}{2a+b} \right)L\]
B) \[\frac{4}{3}\left( \frac{a+b}{2a+3b} \right)L\]
C) \[\frac{3}{2}\left( \frac{2a+b}{3a+b} \right)L\]
D) \[\frac{3}{4}\left( \frac{2a+b}{3a+b} \right)L\]
Correct Answer: D
Solution :
\[{{X}_{cm}}=\frac{\int\limits_{0}^{L}{X\,dM}}{M}\] \[M=aL+\frac{bL}{3}\] \[\int\limits_{0}^{L}{X\,\,dM=}\int\limits_{0}^{L}{\left( aX+\frac{b{{X}^{3}}}{{{l}^{2}}} \right)}dx=\left( \frac{a{{L}^{2}}}{2}+\frac{b{{L}^{2}}}{4} \right)\] \[{{X}_{cm}}=\frac{3L}{4}\,\,\left( \frac{2a+b}{3a+b} \right)\]You need to login to perform this action.
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