question_answer

If the linear density (mass per unit length) of a rod of length 3 m is proportional to\[x,\]where\[x\]is the distance from one end of the rod, the distance of the centre of gravity of the rod from this end is

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. \[{{x}_{CG}}=\frac{\int\limits_{0}^{L}{xdm}}{\int\limits_{0}^{L}{dm}}\] \[{{x}_{CG}}=\frac{\int\limits_{0}^{3}{x(kx\,dx)}}{\int\limits_{0}^{3}{kx}dx}=\frac{\left[ \frac{{{x}^{3}}}{3} \right]_{0}^{3}}{\left[ \frac{{{x}^{2}}}{2} \right]_{0}^{3}}=\frac{\frac{{{(3)}^{2}}}{3}}{\frac{{{(3)}^{3}}}{3}}=2m\]