question_answer

    If linear density of a rod of length 3 m varies as \[\lambda =2+x,\]then the position of the centre of gravity of the rod is

A)  \[\frac{7}{3}m\]                                              

B)  \[\frac{12}{7}m\]

C)  \[\frac{10}{7}m\]                           

D)  \[\frac{9}{7}m\]

Correct Answer: B

Solution :

                Let rod is placed along\[x-\]axis. Mass of element PQ of length dx situated at\[x=x\]is The CM of the element has coordinate\[(x,0,0)\] Therefore,\[x-\]coordinate of CM of the rod will be                 \[{{x}_{CM}}=\frac{\int_{0}^{3}{x\,dm}}{\int_{0}^{3}{dm}}\]                 \[=\frac{\int_{0}^{3}{x(2+x)dx}}{\int_{0}^{3}{(2+x)dx}}\]                 \[=\frac{\int_{0}^{3}{(2x+{{x}^{2}})dx}}{\int_{0}^{3}{(2+x)dx}}\]                 \[=\frac{\left[ \frac{2{{x}^{2}}}{2}+\frac{{{x}^{3}}}{3} \right]_{0}^{3}}{\left[ 2x+\frac{{{x}^{2}}}{2} \right]_{0}^{3}}\]                 \[=\frac{\left[ {{(3)}^{2}}+\frac{{{(3)}^{3}}}{3} \right]}{\left[ 2\times 3+\frac{{{(3)}^{3}}}{2} \right]}=\frac{12}{7}m\]