A) 2.5 m
B) 1 m
C) 1.5m
D) 2m
Correct Answer: D
Solution :
Consider an element of length\[dx\]at a distance \[x\] from end A. Here, mass per unit length\[\lambda \]of rod \[\lambda \propto x\Rightarrow \lambda =kx\] \[\therefore \] \[dm=\lambda dx=kxdx\] Position of centre of gravity of rod from end A. \[{{x}_{CG}}=\frac{\int\limits_{0}^{L}{xdm}}{\int\limits_{0}^{L}{dm}}\] \[{{x}_{CG}}=\frac{\int\limits_{0}^{3}{x(kx\,dx)}}{\int\limits_{0}^{3}{kx}dx}=\frac{\left[ \frac{{{x}^{3}}}{3} \right]_{0}^{3}}{\left[ \frac{{{x}^{2}}}{2} \right]_{0}^{3}}=\frac{\frac{{{(3)}^{2}}}{3}}{\frac{{{(3)}^{3}}}{3}}=2m\]You need to login to perform this action.
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