Four particles of masses m, 2m, 3m and 4m are arranged at the corners of a parallelogram with each side equal to a and one of the angle between two adjacent sides as `60^(@)` The parallelogram lies in the x-y plane with mass m at the origin and 4m on the r-axis. The centre of mass of the arrangement will be located at

Options

A. `((sqrt3)/(2) a , 0.95 a)`
B. `(0.95 a , (sqrt3)/(4) a)`
C. `((3a)/(4) , (a)/(2))`
D. `((a)/(2) , (3a)/(4))`

Answer

B

Explanation

`X_(CM) = (m_(1) x_(1) + m_(2) x_(2) + m_(3) x_(3) + m_(4) x_(4))/(m_(1) + m_(2) + m_(3) + m_(4))`
`= ((m xx 0) + (2m xx (a)/(2)) + (3m xx (3a)/(2)) + (4m xx a))/(m + 2m + 3m + 4m)`
`= (ma + 4.5 ma + 4ma)/(10 m) = (9.5 ma)/(10m) = 0.95` a
`Y_(CM) = (m_(1) y_(1) + m_(2) y_(2) + m_(3) y_(3) + m_(4) y_(4))/(m_(1) + m_(2) + m_(3) + m_(4))`
`(( m xx 0) + (2m xx (a sqrt3)/(2)) + (3m xx (a sqrt3)/(2)) 4 (4m xx0))/(m + 2m + 3m + 4m)`
`= (sqrt3 am + sqrt3 xx 1.5 ma)/(10 m) = (2.5 sqrt3 am)/(10 m) = (sqrt3a)/(4)`
`therefore` Centre of mass is at `(0.95a , (sqrt3a)/(4))`