The object at rest suddenly explodes into three parts with the mass ratio 2: 1: 1. The parts of equal masses move at right angles to each other with equal speed v. The speed of the third part after the explosion will be:

A. 2V
B. $\dfrac{V}{2}$
C. $\dfrac{V}{{\sqrt 2 }}$
D. $\sqrt 2 V$

Hint : A conservation law in physics states that a certain observable property of an independent physical structure does not alter over time. Power conservation, linear momentum conservation, angular momentum conservation, and electric charge conservation are all examples of exact conservation laws. Mass, parity, lepton number, baryon number, strangeness, hypercharge, and other quantities are all subject to approximate conservation laws. Certain groups of physics methods, but not always, preserve these quantities.

Formula Used:
$\sqrt {p_1^2 + p_2^2} = p$
P = momentum of the particle [p = mv]

Complete step-by-step solution:
Until acted upon by an unbalanced force, an object at rest stays at rest and an object in motion stays in motion at the same speed and in the same direction. The ability of all objects to avoid some transition in direction is known as inertia. Until a force acts on it to shift its speed or direction, inertia forces a moving object to remain in place at the same velocity (speed and direction). It also keeps an object at rest from moving.
The law of conservation of momentum states that until an external force is applied, the overall momentum of two or more bodies acting on each other in an independent structure remains unchanged. As a result, neither the development nor the destruction of momentum is possible.
The theory of momentum conservation is a direct result of Newton's third law of motion.
Now let us consider the speed of the third particle be ${v_3}$
Now suppose we apply law of conservation of momentum we have
$\sqrt {p_1^2 + p_2^2} = p$
Taking squares
$ \Rightarrow p_1^2 + p_2^2 = {p^2}$
Let P = m v
${\left( {{m_1}{v_1}} \right)^2} + {\left( {{m_2}{v_2}} \right)^2} = {\left( {{m_3}{v_3}} \right)^2}$
${(m \times v)^2} + {(m \times v)^2} = {\left( {2m \times {v_3}} \right)^2}$
${m^2}{v^2} + {m^2}{v^2} = 4{m^2}v_3^2$
$r{m^2}{v^2} = 4{m^2}v_3^2$
${v_3} = \dfrac{v}{{\sqrt 2 }}$
Hence option C is correct.

Note: Newton's third law states that if object A exerts a force on object B, object B must respond with a force of equal magnitude and in the opposite direction. This rule reflects nature's symmetry: forces often exist in pairs, and one body cannot exert a force on another without undergoing one.