question_answer

A rod of length L has non-uniform linear mass density given by\[\rho \left( x \right)=a+b{{\left( \frac{x}{L} \right)}^{2}}\], where a and b are constants and\[0\underline{<}\,x\,\underline{<L}\]. The value of x for the centre of mass of the rod is at            [JEE MAIN Held on 09-01-2020 Evening]

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 :

\[{{X}_{cm}}=\frac{\int\limits_{0}^{L}{X\,dM}}{M}\] \[M=aL+\frac{bL}{3}\] \[\int\limits_{0}^{L}{X\,\,dM=}\int\limits_{0}^{L}{\left( aX+\frac{b{{X}^{3}}}{{{l}^{2}}} \right)}dx=\left( \frac{a{{L}^{2}}}{2}+\frac{b{{L}^{2}}}{4} \right)\] \[{{X}_{cm}}=\frac{3L}{4}\,\,\left( \frac{2a+b}{3a+b} \right)\]