Explanation:
When a nucleus ruptures, it releases two nuclear parts. Let's assume that the masses of these two parts are m1 and m2, respectively. The velocity ratio of these two parts is given as 2:1. Let's assume that the velocity of the lighter part (m2) is v and the velocity of the heavier part (m1) is 2v.

Conservation of Momentum:
According to the law of conservation of momentum, the total momentum of the two nuclear parts before and after the rupture should be equal.

Before Rupture: The initial momentum of the nucleus is zero as it is at rest.

After Rupture: The total momentum of the two nuclear parts can be calculated as follows:
m1(2v) + m2(v) = 0 (as the initial momentum was zero)

Conservation of Energy:
According to the law of conservation of energy, the total kinetic energy of the two nuclear parts before and after the rupture should be equal.

Before Rupture: The initial kinetic energy of the nucleus is zero as it is at rest.

After Rupture: The total kinetic energy of the two nuclear parts can be calculated as follows:
(1/2)m1(4v^2) + (1/2)m2(v^2) = (1/2)Mv^2 (where M is the mass of the nucleus)

Ratio of Nuclear Size:
Now, we can use the above two equations to find the ratio of the nuclear sizes (nuclear radii) of the two nuclear parts.

m1(2v) + m2(v) = 0
=> m1 = -2m2

Substituting this value of m1 in the second equation, we get:

(1/2)(-8m2v^2) + (1/2)m2(v^2) = (1/2)Mv^2
=> -4m2v^2 + m2v^2 = Mv^2
=> m2/M = 1/3

The ratio of the nuclear sizes (nuclear radii) can be calculated as follows:

Let R1 and R2 be the nuclear radii of m1 and m2, respectively.

We know that the mass and volume of a nucleus are related as follows:
M = (4/3)πR^3ρ
(where ρ is the density of the nucleus)

Therefore, R ∝ M^(1/3)

So, R1/R2 = (m1/M)^(1/3) / (m2/M)^(1/3)
=> R1/R2 = (-2)^(1/3) / 1
=> R1/R2 = 1/(2^(1/3))

Hence, the ratio of the nuclear sizes (nuclear radii) is 1:(2)^(1/3).

Ration of mass by conservation of momentum is 1:2 .then radius is direct proportional to A(mass no.)raised to 1/3.