Correct Answer - D
`a_(cm) = (F_("net"))/("Total mass") = ((0.2)(3)(10))/(1 + 2) = 2ms^(-2)`
Acceleration of `1 kg` w.r.t. ground
`= (0.1)(10) = 1ms^(-2)`
Acceleration of `2 kg` w.r.t. ground
`= ((0.2)(3)(10) - (0.1)(10))/(2) = (5)/(2) ms^(-2)`
`a_(cm) = (m_(1)a_(1) + m_(2)a_(2))/(m_(1) + m_(2)) = ((1)(1) + (2)(5//2))/(1 + 2) = 2ms^(-2)`