A) \[\frac{{{\lambda }_{0}}l+2k{{l}^{2}}}{3(2{{\lambda }_{0}}+kl)}\]
B) \[\frac{3{{\lambda }_{0}}l+2k{{l}^{2}}}{(2{{\lambda }_{0}}+kl)}\]
C) \[\frac{4{{\lambda }_{0}}l+k{{l}^{2}}}{3\,(2{{\lambda }_{0}}+kl)}\]
D) \[\frac{3{{\lambda }_{0}}l+2k{{l}^{2}}}{3\,(2{{\lambda }_{0}}+kl)}\]
Correct Answer: D
Solution :
We have, \[dm=\lambda dx=({{\lambda }_{0}}+kx)dx\] |
\[\therefore \]\[{{X}_{CM}}=\frac{\int{xdm}}{\int{dm}}=\frac{\int\limits_{0}^{l}{x({{\lambda }_{0}}+kx)dx}}{\int\limits_{0}^{l}{({{\lambda }_{0}}+kx)dx}}\] |
\[=\frac{{{\lambda }_{0}}\frac{{{l}^{2}}}{2}+\frac{k{{l}^{3}}}{3}}{{{\lambda }_{0}}l+\frac{k{{l}^{2}}}{2}}=\frac{3{{\lambda }_{0}}l+2k{{l}^{2}}}{3(2{{\lambda }_{0}}+kl)}\] |
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