To solve this problem, we need to apply the conservation of momentum and calculate the kinetic energy generated.$\\$Given:$\\$• Initial mass of the body $= 4m \\$• Initial velocity of the body $= 0$ (since it is at rest)$\\$• Two pieces of mass $m$ each move perpendicular to each other with speed $v. \\$Let's denote:$\\$• Mass of the third piece $= 2m \\$• Velocity of the third piece $= v_3 \\$By conservation of momentum, the total momentum before and after the explosion must be equal.$\\$Initial momentum $= 0 \\$Final momentum $= m \cdot v \hat{i} + m \cdot v \hat{j} + 2m \cdot v_3 \\$Since the initial momentum is zero, the vector sum of the final momenta must also be zero:$\\ m \cdot v \hat{i} + m \cdot v \hat{j} + 2m \cdot v_3 = 0 \\$This implies:$\\ v_3 = -\frac{v}{2} \hat{i} - \frac{v}{2} \hat{j} \\$Now, calculate the kinetic energy of each piece:$\\$• Kinetic energy of each piece of mass $m: \\ K_1 = \frac{1}{2} m v^2 \\ K_2 = \frac{1}{2} m v^2 \\$• Kinetic energy of the third piece of mass $2m: \\ K_3 = \frac{1}{2} (2m) \left(\sqrt{\left(-\frac{v}{2}\right)^2 + \left(-\frac{v}{2}\right)^2}\right)^2 \\$Simplify $K_3: \\ K_3 = \frac{1}{2} (2m) \left(\frac{v}{\sqrt{2}}\right)^2 \\ K_3 = \frac{1}{2} (2m) \left(\frac{v^2}{2}\right) \\ K_3 = \frac{1}{2} m v^2 \\$Total kinetic energy generated:$\\ K_{total} = K_1 + K_2 + K_3 \\ K_{total} = \frac{1}{2} m v^2 + \frac{1}{2} m v^2 + \frac{1}{2} m v^2 \\ K_{total} = \frac{3}{2} m v^2 \\$Therefore, the total kinetic energy generated due to the explosion is $\frac{3}{2} m v^2. \\$This corresponds to Option 2.