A nucleus ruptures into two nuclear parts, which have their velocity ratio equal to `2 : 1`. What will be the ratio of their nuclear size (nuclear radius)?

Options

A. `1:3^(1//2)`
B. `3^(1//2):1`
C. `2^(1//3) :1`
D. `1:2^(1//3)`

Answer

D

Explanation

Nucleus of mass M, at rest, break into two part of masses `m_1 " and " m_2` moving with velocities `V_1 and v_2` respectively.
By the principle of conservation of linear mometum.
`M xx 0 = m_1v_1 + m_2 v_2`
`therefore m_1 v_1 = -m_2v_2`
`therefore v_1/v_2 = - (m_2)/(m_1) = 2/1` ( in magnitude) ...(1)
The nucleus is assumed to be a sphere of radius r and density `rho`
`therefore " it mass " = 4/3 pir^3 rho therefore m prop r^3 "( as " 4/3 pirho` is constant)
`therefore m_2/m_1 = (r_2/r_1)^3 = (v_1/v_2) = (2/1)`
`therefore (r_1/r_2)^3 = 1/2 therefore r_1/r_2 = (1/2)^(1//3) = 1:2^(1//3)`