A shell of mass m is at rest initially.  It explodes into three fragments having mass in the ratio 2 : 2 : 1.  If the fragments having equal mass fly off along mutually perpendicular directions with speed v, the speed of the third (lighter) fragment is   

Given, mass of the shell = m

Ratio of masses of the fragments is 2 : 2 : 1.

Therefore masses of three fragments are

$m_1=\frac{m}{2},m_2=\frac{m}{2}\text{and}{m}_3=\frac{m}{4}$

Now fragments with equal masses i.e. $m_1$ and $m_2$ fly off perpendicularly with speeds $v_1=v_2=v$.  Let the velocity of third fragment be v’.

Applying law of conservation of momentum,

$m_1{v}_1\widehat{i}+m_2{v}_2\widehat{j}+m_3\overrightarrow{v}=0$

$\frac{mv}{2}\widehat{i}+\frac{mv}{2}\widehat{j}+\frac{m}{4}\overrightarrow{v}\text{'}=0\Rightarrow \overrightarrow{v}\text{'}=2v\left(-\widehat{i}-\widehat{j}\right)$

$\left|\overrightarrow{v}\right|=2v\sqrt{{\left(-1\right)}^2+{\left(-1\right)}^2}=2\sqrt{2}v$