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                 \[dm=\lambda dx=(2+x)\,dx\] The COM of the element has coordinates (x, 0,0). Therefore, x-coordinate of COM of the rod will be                 \[{{x}_{COM}}\frac{\int_{0}^{3}{xdm}}{\int_{0}^{3}{dm}}\]                 \[=\frac{\int_{0}^{3}{x(2+x)\,dx}}{\int_{0}^{3}{(2+x)\,dx}}\]                 \[=\frac{\int_{0}^{3}{(2+{{x}^{2}})\,dx}}{\int_{0}^{3}{(2+x)\,dx}}\]                 \[=\frac{\left[ \frac{2{{x}^{2}}}{2}+\frac{{{x}^{3}}}{3} \right]_{0}^{3}}{\left[ 2x+\frac{{{x}^{2}}}{2} \right]_{0}^{3}}\]                 \[=\frac{\left[ {{(3)}^{2}}+\frac{{{(3)}^{3}}}{3} \right]}{\left[ 2\times 3+\frac{{{(3)}^{2}}}{2} \right]}\]                 \[=\frac{9+9}{6+9/2}\]                 \[=\frac{18\times 2}{21}\] \[=\frac{12}{7}\,m\]