The correct option is C 4Eo
Two fragments have same kinetic energy and mass, then their speed must be same.
Let speed of both fragments is v
Both are perpendicular, then consider velocity of first fragment is x− axis direction while velocity of second is y− axis direction.
Therefore, →v1=v^i,→v2=v^j
Third fragment will fly in a direction opposite to the direction of resultant of other two particles. So, the total momentum of the system after explosion also remains zero.
∴ Applying momentum conservation for all the three fragments,
m1→v1+m2→v2+m3→v3=0
→v1+→v2+→v3=0
⇒v^i+v^j+→v3=0
→v3=−(v^i+v^j)⇒|v3|=√2v
Now, kinetic energy of the first and second fragment =E0=12mv2
∴ Total K.E =12mv2+12mv2+12m(√2v)2
=E0+E0+2E0
⇒ Total KE =4E0