Rutherford’s experiments suggested that the size of the nucleus is about

Explanation

Here the sphere of radius R can be considered as a system made of a sphere of radius a and the remaining part.
`therefore` Mass of the sphere of radius R.
`M=(4)/(3)piR^(3)rho[therefore" Density"=("mass")/("volume")]` and Co-ordinate of its centre O is `vec(r_(cm))=(0,0,0)`
`therefore` Mass of the sphere of radius a is
`m_(1)=(4)/(3)pia^(3)rho`
Co-ordinate of its Centre A are `vec(r_(1))=(b,0,0)`
`therefore` Cutting out a sphere of radius a from the sphere of radius R, the mass of the remaining part is
`m_(2)=M-m_(1)`
`=(4)/(3)piR^(3)rho-(4)/(3)pia^(3)rho[becauseM=m_(1)+m_(2)]`
`=(4)/(3)pirho(R^(3)-a^(3)]`
Suppose coordinates of its centre B are
`vec(r_(2))=(x,0,0)`
Now,
`Mvec(r_(cm))=m_(1)vec(r_(1))+m_(2)vec(r_(2))`
`therefore (4)/(3)piR^(3)rho(0,0,0)=(4)/(3)pia^(3)rho(b,0,0)+(4)/(3)pirho(R^(3)-a^(3))(x,0,0)`
Comparing x-coordinates
`0=(4)/(3)pia^(4)rhob+(4)/(3)pirho(R^(3)-a^(3))x`
`therefore (R^(3)-a^(3))x=-a^(3)b[because" Divding by"(4)/(3)pi]`
`therefore x=-(a^(3)b)/((R^(3)-a^(3)))therefore vec(r_(2))=(x,0,0)`
`therefore vec(r_(2))=(-(a^(3)b)/((R^(3)-a^(3))),0,0)` unit