Question

An infinite number of bricks are placed one over the other as shown in the figure. Each succeeding brick having half the length and breadth of its preceding brick and the mass of each succeeding brick being ${\frac{1}{4}}^{\text{th}}$ of the preceding one. Take $O$ as the origin. The $x-$coordinate of centre of mass of the system of bricks is:

• A. $-\frac{a}{7}$
• B. $\frac{3a}{7}$
• C. $-\frac{3a}{7}$
• D. $-\frac{2a}{7}$

Answer 1

The correct option is C $-\frac{3a}{7}$

Centre of mass of each block will be at its mid point.
For mass $m_1=$$M$, $x_1=\frac{-a}{2}$
For mass $m_2=$$\frac{M}{4}$, $x_2=\frac{-a}{4}$ and so on.

Position of centre of mass is given by :
$X_{com}=\frac{m_1{x}_1+m_2{x}_2+...........\infty }{m_1+m_2+......\infty }$
$=\frac{M\left(\frac{-a}{2}\right)+\frac{M}{4}(-\frac{a}{4})+\frac{M}{16}(-\frac{a}{8})+.......}{M+\frac{M}{4}+\frac{M}{16}+......}$
$=\frac{-\frac{a}{2}[1+\frac{1}{8}+\frac{1}{64}+......]}{1+\frac{1}{4}+\frac{1}{16}+......}$

Sum of an infinite GP is given by : $s_{\infty }=\frac{a}{1-r}$
$\Rightarrow X_{com}$$=\frac{\frac{-a}{2}\left(\frac{8}{7}\right)}{\frac{4}{3}}=\frac{-3a}{7}$