question_answer

A uniform thin rod AB of length L has linear mass density \[\mu (x)=a+\frac{bx}{L},\]where x is measured from A. If the CM of the rod lies at a distance of \[\left( \frac{7}{12}L \right)\]from A, then a and b are related as : [JEE Main Online Paper (Held On 11 April 2015)]  

A)  a = b

B)  a = 2b

C)  2a = b

D)  3a = 2b

Correct Answer: C

Solution :

  \[{{X}_{COM}}=\frac{\int\limits_{0}^{L}{\left( \mu dx \right)x}}{\int\limits_{0}^{L}{\mu dx}}\] \[\frac{7}{12}L=\frac{\int\limits_{0}^{L}{\left( ax+\frac{b{{x}^{2}}}{L} \right)dx}}{\int\limits_{0}^{L}{\left( a+\frac{bx}{L} \right)dx}}\] \[\frac{7}{12}L=\frac{\left( \frac{b{{L}^{3}}}{3L}+\frac{a{{L}^{2}}}{2} \right)}{\left( aL+\frac{bL}{2} \right)}\] \[\frac{7}{12}=\frac{\frac{a}{2}+\frac{b}{3}}{a+\frac{b}{2}}\] \[\therefore \]\[\] \[\therefore \,\,\,\,2a=b\]