question_answer

If the linear density (mass per unit length) of a rod of length 3m 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) 2.5m

B) 1m 

C) 1.5m

D) 2m

Correct Answer: D

Solution :

[d] Consider an element of length dx at a distance x from end A. Here, mass per unit length \[\lambda \] of rod \[\begin{align}   & \lambda \propto x\Rightarrow \lambda =kx \\  & \therefore \,\,dm=\lambda dx=kxdx \\ \end{align}\] Position of centre of gravity of rod from end A. \[{{x}_{CG}}=\frac{\int_{0}^{L}{xdm}}{\int_{0}^{L}{dm}}\] \[\therefore \,\,{{x}_{CG}}=\frac{\int_{0}^{3}{x(kx\,dx)}}{\int_{0}^{3}{kxdx}}=\frac{{{\left[ \frac{{{x}^{3}}}{3} \right]}^{3}}}{\left[ \frac{{{x}^{2}}}{2} \right]_{0}^{3}}=\frac{\frac{{{(3)}^{3}}}{3}}{\frac{{{(3)}^{3}}}{2}}=2m\]