An infinite no of bricks are placed one over the other as shown in the figure. Each succeeding brick having half the length and breadth of its precceeding brick and the mass of each succeeding bricks `1//4^(th)` of the preceding one, take 'O' as the origin, the centre of mass of the system of bricks is at

Options

A. `-a/7`
B. `-(5a)/7`
C. `-(3a)/7`
D. `-(2a)/7`

Answer

3

Explanation

`X_(cm^(2))=(M_(1)x_(1)+M_(2)x_(2)+........)/(M_(1)+M_(2))`
`=(M((-a)/2)+M/4((-a)/4)+M/16((-a)/8)+......)/(M+M/4+M/16+M/64.........)`
`([(Ma)/2+(Ma)/10+(Ma)/128+.....])/([M+M/4+M/16+......])`
`(M[(a//2)/(1-(1//8))])/([(1/(1-(1//4))])=(-3a)/7`