Question

If linear density of a rod of length 3 m varies as $\lambda =1+x$, then the position of the centre of mass of the rod is

• A. $\frac{7}{3}\text{m}$
• B. $\frac{9}{5}\text{m}$
• C. $\frac{7}{2}\text{m}$
• D. $\frac{9}{2}\text{m}$

Answer 1

The correct option is B $\frac{9}{5}\text{m}$

Linear density of the rod varies with distance
$\frac{dm}{dx}=\lambda \text{[Given]}\therefore dm=\lambda dx$


Position of centre of mass $x_{CM}=\frac{\int dm\times x}{\int dm}$
$\frac{{\int }_3^0\left(\lambda dx\right)\times x}{{\int }_0^3\lambda dx}=\frac{{\int }_0^3(1+x)\times xdx}{{\int }_0^3(1+x)dx}=\frac{{[0.5x^2+\frac{x^3}{3}]}_0^3}{{[x+\frac{x^2}{2}]}_0^3}$
$=\frac{4.5+9}{3+\frac{9}{2}}=\frac{9}{5}\text{m}$
Hence, the correct answer is option (b).