A bomb of mass 4m at rest explodes into 3 pieces of masses m , m ,...

By the law of conservation of linear momentum, the initial momentum of the bomb is zero. Thus, the vector sum of the momenta of the three fragments must be zero.

Let the two fragments of mass $m$ move along the x and y axes. Their momenta are $\vec{p}_1 = m v \hat{i}$ and $\vec{p}_2 = m v \hat{j}$.

The momentum of the heaviest fragment (mass $2m$) is $\vec{p}_3 = -(\vec{p}_1 + \vec{p}_2) = -m v \hat{i} - m v \hat{j}$.

Magnitude of momentum of the heaviest fragment: $p_3 = \sqrt{(mv)^2 + (mv)^2} = \sqrt{2} m v$. (Matches II)

Speed of the heaviest fragment: $v_3 = \dfrac{p_3}{2m} = \dfrac{\sqrt{2} m v}{2m} = \dfrac{v}{\sqrt{2}}$. (Matches I)

Kinetic energy of the heaviest fragment: $K_3 = \dfrac{p_3^2}{2(2m)} = \dfrac{2 m^2 v^2}{4m} = \dfrac{1}{2} m v^2$. (Matches IV)

Total kinetic energy generated: $K_{total} = K_1 + K_2 + K_3 = \dfrac{1}{2} m v^2 + \dfrac{1}{2} m v^2 + \dfrac{1}{2} m v^2 = \dfrac{3}{2} m v^2$. (Matches III)

Therefore, the correct matching is (A)-(II), (B)-(I), (C)-(IV), (D)-(III).

Answer: (A)-(II), (B)-(I), (C)-(IV), (D)-(III)