Three identical spheres each of diameter `2 sqrt(3)` m are kept on a horizontal surface such that each sphere touches the other two spheres. If one of the sphere is removed, then the shift in the position of the centre of mass of the system is

Options

A. 12 m
B. 1 m
C. 2 m
D. `3/2m`

Answer

B

Explanation

The centre of mass of the spheres is given by
`x_(cm) = (m_(1) x_(1) + m_(2) x_(2) + m_(3) x_(3))/(m_(1) +m^(2)+ m^(3))`

As the spheres are identical,
`rArr x_(cm)=m/(3m)(0 + 2 sqrt(3) + sqrt(3))`
= `(3sqrt(3))/3 = sqrt(3) `
Similarly, `y_(cm) = m/(3m) (y_(1) + y_(2) + y_(3))`
` = (0+0+3)/3 = 1`
So, the centre of mass, `C_(cm) = (sqrt(3) , 1)` If one sphere is removed (say C) , then
`x_(cm) = (0+2sqrt(3) )/2 = sqrt(3)`
`y _(cm) = 0`

So, `C_(cm) = (sqrt(3), 0)`
Hence, the centre of mass shifted by 1 m in - y direction. The correct option is (b) .