To solve this problem, we will use the principle of conservation of momentum.$\\$Given:$\\$• The shell is initially at rest, so its initial momentum is zero.$\\$• The shell explodes into three fragments with mass ratio $2:2:1. \\$Let's denote:$\\$• Total mass of the shell as $m. \\$• Mass of each of the two equal fragments as $m_1 = \frac{2m}{5}. \\$• Mass of the third (lighter) fragment as $m_2 = \frac{m}{5}. \\$• Speed of the two equal fragments as $v. \\$• Speed of the third fragment as $v_2. \\$Since the initial momentum is zero, the total final momentum must also be zero.$\\$The two equal fragments fly off along mutually perpendicular directions.$\\$Let's assume one fragment moves along the x-axis and the other along the y-axis.$\\$The momentum of the first fragment is $m_1 v$ in the x-direction.$\\$The momentum of the second fragment is $m_1 v$ in the y-direction.$\\$The momentum of the third fragment must balance these in the opposite direction.$\\$Using vector addition, the magnitude of the resultant momentum of the two fragments is:$\\ p_{resultant} = \sqrt{(m_1 v)^2 + (m_1 v)^2} = \sqrt{2} m_1 v \\$This resultant momentum must be equal and opposite to the momentum of the third fragment:$\\ m_2 v_2 = \sqrt{2} m_1 v \\$Substitute the mass values:$\\ \frac{m}{5} v_2 = \sqrt{2} \left(\frac{2m}{5}\right) v \\$Simplify:$\\ v_2 = \sqrt{2} \cdot 2v \\ v_2 = 2\sqrt{2} v \\$Therefore, the speed of the third (lighter) fragment is $2\sqrt{2} v. \\$This corresponds to Option 3.