A rod of length is $3 \mathrm{~m}$ and its mass acting per unit length is directly proportional to distance $x$ from one of its end, then its centre of gravity from that end will be at:

• A

  $1.5 \mathrm{~m}$
• Correct answer
  B

  $2 \mathrm{~m}$
• C

  $2.5 \mathrm{~m}$
• D

  $3.0 \mathrm{~m}$

Solution

Here $r=k x$ where $k=$ constant mass of small element of $d x$ length is $\begin{aligned} & d m=k x \cdot d x \\ & \mathrm{X}_{\mathrm{CM}}=\frac{\int x d m}{\int d m}=\frac{\int_0^3 x(x d x)}{\int_0^3 x d x} \\ &=\frac{\left[\frac{x^3}{3}\right]_0^3}{\left[\frac{x^2}{2}\right]_0^3}=\frac{\frac{27}{3}}{\frac{9}{2}}=2 \end{aligned}$

Hence, the correct answer is option: 'B'.